calculus - Prove the relation involving derivative of inverse of a function

I want to prove the following result:

$${f^{-1}}'(x)=\frac{1}{f'({f^{-1}}(x))}$$

Is simple application of chain rule a valid proof of it?

i.e. $$f({f^{-1}}(x))=x \implies \frac{df({f^{-1}}(x))}{dx}=1$$ and hence the result. Or is this not a standard proof? Is there additional conditions necessary, expect of course that the function is bijective, or that the inverse exists.

Answer

The question is: Why is $f^{-1}$ differentiable? In the theorem you probably learned in a lecture this isn't assumed. However if you assume that $f^{-1}$ and $f$ are differentiable at and all the expressions make sense $f^{-1}$ should be a function and $f'\neq 0$ (both at least locally).

But nevertheless, with the chainrule the formula is easy to remember.