The discussion here has been quite interesting! I wrote about Leibniz's notation in my Bachelor's Thesis in 2010 reading through major parts of Bos's 1974 PhD on higher order differentials in the Leibnizian calculus. I believe Bos is wrong at one point. Assuming one variable in the by Bos so-called arithmetic progression is never necessary - only convenient! I will answer to that below.
Leibniz's differentials
Leibniz developed his differentials, at first, from a geometrical intuition - although he reconsidered the actuality of this idea time and again. In my words, this idea can be very briefly summarized as:
A curve can be thought of as a polygon with infinitely many infinitely small sides $ds$. Each $ds$ is an infinitesimally small straight line segment being a part of the curve and (paradoxically) tangent to it at the same time. Gathering the $ds$ to one straight line segment $s=\int ds$ this will constitute the length of the curve. Expressing such a curve by a geometrical relation between coordinate line segments $x$ and $y$ one could consider each $ds$ as hypotenuse of a right triangle with legs $dx$ and $dy$ so that $dx^2+dy^2=ds^2$.
This is only to say that $dx,dy$ and $ds$ was thought of as geometrical and mutually dependend entities - never considered just numbers like we allow functions to be today.
Just to stress how geometrical: the function nowadays expressed by the formula $f(x)=x^2$ would be something like $a\cdot y=x\cdot x$ where $a,y$ and $x$ where all considered line segments so that both sides of the equation would constitute an area in Leibniz's time.
The level curve example
In the fractions $\frac{\partial f}{dx}$ and $\frac{\partial f}{dy}$ the $\partial f$'s in the two fractions are unrelated because:
- We do not have $\partial f,\partial x$ and $\partial y$ mutually dependend geometrical entities due to the reason you already gave that the first $\partial f$ is the change in $f$ when you move in the $x$-direction by the vector $(dx,0)$ whereas the second $\partial f$ corresponds to moving by the vector $(0,dy)$. So they are unequal although infinitesimally small ...
- Even if we had some $df$ mutually dependend to $dx$ and $dy$ this would naturally have to be the change in $f$ when you travel the vector $(dx,dy)$ and thus different from the $\partial f$'s described before.
The chain rule example
Since we consider higher order differentials the work of Bos is relavant here: Had there been such thing as a derivative $z=\frac{dy}{dv}$ in Leibniz's time, the differential of that should read
$$
dz=d\frac{dy}{dv}=\frac{dy+ddy}{dv+ddv}-\frac{dy}{dv}=\frac{dv\ ddy-dy\ ddv}{dv(dv+ddv)}
$$
Now, since $ddv$ is infinitesimally small compared to $dv$ we may skip $ddv$ in the bracket and simply write $dv$ instead of $(dv+ddv)$. Therefore we have
$$
\frac{dz}{dv}=\frac{dv\ ddy-dy\ ddv}{dv^3}=\frac{ddy}{dv^2}-\frac{dy\ ddv}{dv^3}
$$
Note that $ddy$ can also be written as $d^2 y$. So the second order derivative of $y$ with respect to $v$ equals $\frac{d^2 y}{dv^2}$ minus some weird fraction $\frac{dy\ d^2 v}{dv^3}$ which can only be disregarded if it is zero. This only happens if either $dy=0$ or $d^2 v=0$. Choosing $d^2 v$ identical zero does the trick and renders $dv$ constant.
Suppose now that $d^2 v\equiv 0$. Then for the example $y=u=v^2$ we see that $du=2v\ dv$ and furthermore $ddu=2v\ ddv+2\ dv^2=2\ dv^2$ where the last equality is due to our choice that $ddv$ is identical zero. Therefore we see that the derivative of $w=\frac{dy}{du}$ will be given as
$$
\frac{dw}{du}=\frac{d^2 y}{du^2}-\frac{dy\ ddu}{du^3}
$$
where the last fraction is far from being zero as it may be rewritten - noting that $y=u\implies dy=du$ and that $\frac{dv}{du}=\frac{1}{2v}$ - to obtain
$$
\require{cancel}
\frac{\cancel{dy}\ ddu}{\cancel{du}\cdot du^2}=\frac{2\ dv^2}{du^2}=\frac{1}{2v^2}
$$
This shows that assuming $\frac{d^2 y}{dv^2}$ to be the second order derivative of $y=v^2$ with respect to $v$ in the modern sense makes $\frac{d^2 y}{du^2}$ differ by $\frac{1}{2v^2}$ from being the second order derivative of $y=u$ with respect to $u$. Now since we know that $y=u$ we have $w=\frac{dy}{du}=1$ and thus $\frac{dw}{du}=0$. Therefore we must have
$$
\frac{d^2 y}{du^2}-\frac{1}{2v^2}=0
$$
in this case showing that $\frac{d^2 y}{du^2}=\frac{1}{2v^2}$. So with the choice $y=u=v^2$ and $ddv\equiv 0$ the equation
$$
\frac{d^2 y}{du^2}\cdot\left(\frac{du}{dv}\right)^2=\frac{d^2 y}{dv^2}
$$
may be successfully checked applying that $\frac{du}{dv}=2v$ since we then have
$$
\frac{1}{2v^2}\cdot(2v)^2=2
$$
which is actually true. This is NOT a coincidence!
Conclusion
The above calculations show that Julian Rosen's very appealing example of failure in the method of the Leibnizian calculus seems to be a misunderstanding about what is meant by the notions of $d^2 y$ and the hidden, but important, additional variables $ddv$ and $ddu$. This provides specific details regarding the comments given by user72694 below the answer from Julian.
However, proving that Leibniz's notation will never produce false conclusions when handled correctly is a whole different story. This is supposedly what Robinson managed to do, but I must admit that I have not read and understood that theory myself.
My Bachelor's thesis focused mainly on understanding how the method was applied by Leibniz and his contemporaries. I have often times thought about the foundations, but mainly from a 17th century perspective.
Comment on Bos's work
On page 31 in his thesis, Bos argues that the limit
$$
\lim_{h_1,h_2\rightarrow 0}\frac{[f(x+h_1+h_2)-f(x+h_1)]-[f(x+h_1)-f(x)]}{h_1 h_2}
$$
only exists if $h_1=h_2$ which then makes this limit equal $f''(x)$. But that is in fact not entirely true. The $x$-differences $h_1$ and $h_2$ need not be equal. It suffices for them to converge to being equal which is a subtle, but important, variation of the setup. We must demand that $h_1$ and $h_2$ converge to zero in a mutually dependend fashion so that
$$
\lim_{h_1,h_2\rightarrow 0}\frac{h_2}{h_1}=1
$$
With this setup the limit of the large fraction from before may still exist, but need not equal $f''(x)$. Since $h_1,h_2$ play the role of $dx$'s this is equivalent to allowing $dx_1\neq dx_2$ so that $ddx=dx_2-dx_1\neq 0$ although being infinitely smaller than the $dx$'s.
This means that it is in fact possible to imitate the historical notion of $dx$ being constant (and thereby $x$ in arithmetic progression) directly by modern limits.
Extras regarding the OP's answer
You are quite right that the differentials can be succesfully manipulated into the equation
$$
\frac{d^2}{dv^2}\big(y(u(v))\big)=y''(u(v))\cdot u'(v)^2+y'(u(v))\cdot u''(v)
$$
under the assumption that $ddv\equiv 0$.
There is, however, a more obvious and even less restrictive choice to leave the progressions of all three variables $u,v$ and $y$ unspecified, and yet to connect the notation in a meaningful way to modern standards:
Introduce a fourth variable $t$ in arithmetic progression (i.e. $ddt\equiv 0$). One could think of it as a time variable so that $u(t),v(t)$ and $y(t)$ are coordinate functions of some vector valued function. Then Julian Rosen's equation can be directly transformed to
$$
\frac{\left(\frac{d^2 y}{dt^2}\right)}{\left(\frac{du^2}{dt^2}\right)}\cdot\left(\frac{\left(\frac{du}{dt}\right)}{\left(\frac{dv}{dt}\right)}\right)^2=\frac{\left(\frac{d^2 y}{dt^2}\right)}{\left(\frac{dv^2}{dt^2}\right)}
$$
and since $dt$ is in arithmetic progression $y''(t)=\frac{d^2 y}{dt^2}$ so that this may be written in modern notation as
$$
\frac{y''(t)}{u'(t)^2}\cdot\left(\frac{u'(t)}{v'(t)}\right)^2=\frac{y''(t)}{v'(t)^2}
$$
which is easily verified to be correct. This is probably the simplest account, but it only uses but does not give a very clear example of the necessity of choosing the progression of the variables. I think my first account did that better.
$$\sqrt[dx]{dy} = \sqrt[du]{dy}^{du \over dx}$$
The real question though isn't whether they are consistent, but rather which rules are the consistent ones and which aren't. Btw, I wonder why we use the notation ${d^2 y \over dx^2}$ rather than ${d^2 y \over d^2 x}$?
– DanielV Apr 25 '14 at 09:41