Clarification about the triple product identity for partial derivatives

Question

I'm having a hard time understanding why \begin{equation}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=-1 \tag{1}\label{1} \end{equation} Wikipedia provides this derivation. I have two problems with it.
The proof starts by stating that there is a function f such that $f(x,y,z)=0$ and that $z$ can be made a function of $x,y$. Furthermore it states that there can be found a curve, along which $dz=0$ and $y$ is a function of $x$, such that we can then write the differential of $z$ in terms of the differential of $x$ as $$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}dx$$
The rest follows naturally from setting $dz=0$ and multiplying some partial derivatives by their inverses.
I have two problems with this proof

1.Chain rule

The first one is that since \begin{equation} \tag{2} \label{2} \frac{\partial z}{\partial x}=-\frac{\partial z}{\partial y}\frac{\partial y}{\partial x} \end{equation} That would mean that, by chain rule, $$\frac{\partial z}{\partial x}=-\frac{\partial z}{\partial x}$$ which would imply this partial derivative to be zero. However if this is true, \ref{1} yields $0=-1$. Is the chain rule not valid in this case? If so, why?

2.Inverse of the partials
The second is that, while applying the last step, it is implied that we obtain \ref{1} by multiplying by the inverse of the righthand-side in \ref{2}. I thought the relationship $$\frac{\partial y}{\partial x}=\frac{1}{\frac{\partial x}{\partial y}}$$ was in general not true, as pointed out in this post. Is it true in this case? And if so, why is that?
Also, if that really is the case, then using the chain rule again yields, from \ref{1}, $$\frac{\partial x}{\partial z}\frac{\partial z}{\partial x}=-1 \iff 1=-1$$ What am I doing wrong?

Answer 1

At least as a place-holder, in line with @GerryMyerson's apt comment:

Yes, there is an appealing heuristic that suggests that ${\partial z\over \partial x}={\partial z\over \partial y}{\partial y\over \partial x}$, ... and such things.

In a different universe, it might not matter that these named variables were related by $f(x,y,z)=0$ or $z=f(x,y)$ or some other relation. But, in our universe, this does have some relevance.

Even in a simpler situation, $f(x,y)=0$, whether or not we rename $f$ to $z$, a person might imagine that (via some sort of implicit function theorem, making $y$ a function of $x$) ${\partial y\over \partial x}={\partial f\over \partial x}/{\partial f\over \partial y}$... but that's off by a sign!?!?! :)

Careful application of the chain rule corrects the sign. :)

EDIT: When $y$ is (locally) defined as a function of $x$ by a relation $f(x,y)=0$, differentiating this with respect to $x$ gives $$ 0 \;=\; f_1(x,y)\cdot {dx\over dx} + f_2(x,y)\cdot {dy\over dx} \;=\; f_1(x,y)+f_2(x,y){dy\over dx} $$ where $f_i$ is the partial derivative of $f$ with respect to the $i$-th argument. This gives $$ {dy\over dx} \;=\; -{f_1(x,y)\over f_2(x,y)} $$ If we somewhat-abuse notation by thinking that $f_1=f_x$ and $f_2=f_y$, then this would be $$ {dy\over dx} \;=\; -{{\partial f\over \partial x}\over {\partial f\over \partial y}} $$ which does not give the expected heuristic outcome, being off by a sign. :)

Answer 2

It has come to me with all your help that my confusion was only a matter of damned notation and that in fact the chain rule is not broken in the implicit function theorem. It all boils down to what the original wikipedia article calls partial derivative and the way I also thought of it. For me, partial differentiation is always when only one parameter is free to move. It is the natural way of defining it in real analysis. Everything else is just a derivative of the composition of a function with a parametrization (what the physicists like to call the "total derivative"). And that is the derivative that is being used and that eluded me. So as an exercise and to check that I have understood everything, I will answer my own post and "correct" the enunciation of the triple product.
Let $f=f(x,y,z)$, $z=z(x,y)$ and $y=y(x)$. Consider the differential of z, $$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$ This is the definition of the differential. But now , when considering the differential of y, we shouldn't write $dy=\frac{\partial y}{\partial x}dx$, but $dy=\frac{d y}{d x}dx$. This is the notation I was used to and a notation which indicates that not only x is being free to vary: we are taking the derivative of y=y(x,z) composed with z=z(x). We can even see the relationship of this derivative with $\frac{\partial y}{\partial x}$, applying the chain rule. $\frac{dy}{dx}=\frac{\partial y}{\partial x}+\frac{\partial y}{\partial z}\frac{dz}{dx}$. This makes sense. y is a function of x, but also of z (were it not the case, $\frac{\partial y}{\partial z}$ in the triple product wouldn't even make sense). Then, we write $$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}\frac{dy}{dx}dx$$ Moving now along a path where $dz=0$ $$\frac{\partial z}{\partial x}=-\frac{\partial z}{\partial y}\frac{dy}{dx} \tag{3} \label{3}$$ Now, to address my (1). Of course the right hand side is not $\frac{\partial z}{\partial x}$.$\frac{\partial z}{\partial x}$ would be $\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}$, where the difference has been discussed earlier.
To adress (2), the relationship is in general not true, but, in this case differentiating the function $f$ with respect to $y$ and $x$ and comparing the two will show that $\frac{\partial z}{\partial x}\frac{\partial x}{\partial z}=1$ and so we write the "corrected" formula, $$-1=\frac{\partial x}{\partial z}\frac{\partial z}{\partial y}\frac{dy}{dx}$$ I hate thermodynamics