More Deapth for Question 1:
Andr'e noted that the idea of multiplying by dx is "mathematically dubious", and in reality there are many situations where this technique will get you into trouble if you apply it haphazardly without understanding what's going on in the background. To the end of providing that background I'm going to go throug the solution process of your first example with as much detail about each step as possible.
$$
{{1}\over{f(x)}}{{df}\over{dx}} = {{x^3}}
$$
casting this to an integration problem:
$$
\int {{f'(x)} \over {f(x)}}dx = \int {{x^3}} dx
$$
now here's where things change from simply thinking about "multiplying" by a differential. in stead of using that (as mentioned before) problematic through process we'll use integration by substitution, but for a very special case.
What this process really means geometrically is using the infinitesimal breakdown of distances along the function itself as the measure for our calculated area (in stead of the distance along the real line).
Literally stating:
$$
du = \lim_{h \to 0} {{f(x)-f(x+h) \over {h}}dx}
$$
which when plugged into the Riemann integral becomes:
$$
{\int {{f'(x)} \over {f(x)}}dx} = \lim_{max(x_i - x_0) \to 0}\sum_{i=1}^{n}{{f(x_i)-f(x_0)}\over {f(x_i)(x_i-x_0)}}{(x_i - x_0))}
$$
$$
= \lim_{max(x_i - x_0) \to 0}\sum_{i=1}^{n}{{1}\over {f(x_i)}}{(f(x_i)-f(x_0))}
$$
by taking the observation that our limit can be changed to:
$$
= \lim_{max(f(x_i) - f(x_0)) \to 0}\sum_{i=1}^{n}{{1}\over {f(x_i)}}{(f(x_i)-f(x_0))}
$$
wihtout altering the meaning (this result is actually non-trivial but I won't get into that here) we have demonstrated that the integrals are equivalent.
This then results in the integral:
$$
\int {{1} \over {f(x)}}df(x) = \int {{x^3}} dx
$$
As a digression; often in calculus classes you're taught this using a variable as a stand in for your function, such as $u = f(x)$, and then do something like $du = f'(x)dx$ to make things easier on your sensibilites for the time (as using a function as a variable isn't usually as easy to grasp, and comes along with other issues). In this case I'll follow both concepts through so you can see how they connect.
The subsitution method with $u = f(x)$ and $du = f'(x)dx$ would then look like:
$$
\int {{f'(x)dx} \over {u}} = \int {{du} \over {u}} = \int {{x^3}} dx
$$
and from here we either get (if I combine the constants for simplicity):
$$
ln(|f(x)|+c = {{x^4} \over {4}}
$$
or with the substitution:
$$
ln(|u|+c = {{x^4} \over {4}}
$$
where we need to put the f(x) back in for u (since they're equal) and get:
$$
ln(|f(x)|+c = {{x^4} \over {4}}
$$
either way.
From here it's just cancellations and algebra until you get your final result.
There are some important things of note here for a general problem of the type:
$$
g(f(x))f'(x) = h(x)
$$
Since the integral is actually some $ \int g(f(x)) df(x)$ if the integral of the function g fails to exist on the region of concern the solution cannot be found using this method. Somtimes by simply "cancelling" differentials, or "multiplying through" by them this gets obfuscated.
