TITLE: Why do most of the operations treating $dy/dx$ as a ratio works?
QUESTION [9 upvotes]: We all know that since $\textrm{d}y$ and $\textrm{d}x$ are not numbers, we can't treat $\frac{dy}{dx}$ as a ratio. However, in various operations, this treatment seems to work. I've picked a few examples:

Chain Rule:
$$\frac{du}{dx} = \frac{du}{dy} \frac{dy}{dx}$$
Variable separation:
$$ \frac{dy}{dx} = f(x)h(y)$$
$$\frac{dy}{h(y)} = f(x)dx$$
Integral of $\frac{dy}{dx}$:
$$ \int \frac{dy}{dx}dx = \int dy = y$$
And there is also some examples that this treatment does not work:
$$\frac{dy}{dx} = - \frac{\partial F / \partial y}{\partial F / \partial x}$$
Where the minus sign shows that this is a counterexample. Now, if this isn't correct, why does it works for most cases? And, if it works for most cases, why there are a few cases that it won't work? Thanks.

References
1) Answer from asmeurer in Is $\frac{dy}{dx}$ not a ratio?

REPLY [2 votes]: $\mathrm{d}x$ and $\mathrm{d}y$ aren't numbers, but they are differentials. And differentials can be multiplied by scalars. 
And if $\mathrm{d}x$ is nonsingular ($x$ being a so-called "independent variable" guarantees this), then if  $z \, \mathrm{d}x = \mathrm{d} y $ has any solutions for $z$, it has a unique solution.
And thus, $z$ really is the ratio of $\mathrm{d}y$ to $\mathrm{d}x$, and it makes sense to say $z = \frac{\mathrm{d}y}{\mathrm{d}x}$.

The usual notation for partial differentiation is... bad. Partial derivatives aren't like ratios at all. Don't treat them like ratios.
More accurately, partial derivatives only become ratios after making a bunch of modifications to the formula being differentiated... and the notation doesn't even tell you what precisely those modifications are supposed to be; you have to infer from context.
For example, if the differential of some bivariate formula is
$$ \mathrm{d}F(x,y) = g(x,y) \mathrm{d} x + h(x,y) \mathrm{d} y $$
then what the notation $\partial F(x,y) / \partial x$ means is:

First infer that one means to single out $x$ and $y$ as "independent" variables.
Restrict all the "independent" variables other than $x$ to be constant — in particular, you have $\mathrm{d}y = 0$.
Simplify the equation to $\mathrm{d}F(x,y) = g(x,y) \mathrm{d} x$.
Now, there really is a ratio between the two differentials. Define $\partial F(x,y)/\partial x = g(x,y)$ to be that ratio.

Actually, rather than the last two steps, it's more accurate to say that $\partial / \partial x$ means to get to additionally apply to $\mathrm{d}F(x,y)$ the transformation $\mathrm{d}x \mapsto 1$ alongside $\mathrm{d}y \mapsto 0$, but I wanted to give a description that at most resembled a ratio.