When can't $dy/dx$ be used as a ratio/fraction?

For functions of one variable, I have never seen a problem, and wouldn't hesitate to treat them as fractions (multiplicatively). However, suppose you have $F(x,y)$ which implicitly defines a function $y=f(x)$, then

$\dfrac{dy}{dx} = -\dfrac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$

If you just straight cancel as fractions, you'd get the wrong sign.

Edit: I just thought; there is a problem with the notation used which makes this error possible. The $\partial F$'s are different! One is given constant $x$ and the other constant $y$, hence they shouldn't necessarily cancel as they do in fractions. I guess it's rather pretty how they do manage to cancel to give a $-1$, but this particular case as just one instance, it's entirely possible for one symbol to represent different things in the same expression, so one would have to be much more careful about cancelling terms.

I do not know very much. I am still studying like you. One thing I know hope this would help and I might be wrong.

If we consider $\dfrac{dy}{dx}$ as a fraction and say $\dfrac{dy}{dx} = \dfrac{1}{\dfrac{dx}{dy}}$ because it is a fraction then there is a problem.
Consider $y=f(x)=x^0$. Now $\dfrac{dy}{dx}=0$ but $\dfrac{dx}{dy}$ is undefined so we see that $\dfrac{dy}{dx} \neq \dfrac{1}{\dfrac{dx}{dy}}$.

Currently I am reading this book for calculus. The author has explained chain rule very good so you might like it.

I had somewhere read that for chain rule to work $x$ and $y$ should be expressible in form $\phi(x,y)=0$. - I might be wrong.

I had also read somewhere that $y=f(x)$ should be a bijective function. $-$ Again I might be wrong here.