Given a function $f(x)$, let $g=f^{-1}(x)$. Then define the composite function $f(g(x))$.
Then, by the Chain Rule, $(f(g(x)))'=f'(g(x)) g'(x)$.
However, since $f(x)$ and $g(x)$ are inverses, $f(g(x))=x$ and $(f(g(x)))'=(x)'=1$. So $1=f'(g(x)) g'(x)$.
If you solve for $g'(x)$, you get $$\frac{1}{f'(g(x))}$$ which is the derivative of an inverse function.
QED
Is the above proof correct and if so, can it be considered mathematically rigorous?
Suppose that $f$ is differentiable on $\Bbb{R}$, and $g(x)=f^{-1}(x)$ for all $x\in\DeclareMathOperator{\ran}{ran}\ran(f)$. By definition of an inverse function, $f(g(x))=x$. If we take the derivative of both sides, we get $$ f'(g(x)) \cdot g'(x)=1 $$ and rearranging this becomes $$ g'(x)=\frac{1}{f'(g(x))} $$ or $$ (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \, . $$ The problem with calling the above working a proof is that we haven't justified why $$ (f \circ g)'(x)=f'(g(x)) \cdot g'(x) \, . $$ You might be tempted to say "well, that's just the chain rule", but be careful. The chain rule is not simply the formula $$ (f \circ g)'(x) = f'(g(x)) \cdot g(x) \, , $$ but rather the following statement:
If $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$, then $f \circ g$ is differentiable at $x$, and $$ (f \circ g)'(x)=f'(g(x)) \cdot g'(x) $$
So have the hypotheses of the chain rule been met? Since $f$ is a differentiable function, it is certainly true that $f$ is differentiable at $g(x)$. But we haven't shown that $g=f^{-1}$ is differentiable at $x$, and so we are unable to call the above working a proof. Just to give you an idea for how things might go wrong, consider what would happen if $f'(f^{-1}(x))=0$. Hence, all that we have shown above is that if $f^{-1}$ is differentiable at $x$, then $$ (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \, . $$ However, with more work we can prove the following theorem:
Let $f$ be a continuous one-to-one function defined on a certain interval, and suppose that $f$ is differentiable at $f^{-1}(x)$, with derivative $f'(f^{-1}(x))\neq 0$. Then, $f^{-1}$ is differentiable at $x$, and $$ (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \, . $$
Because of this theorem, it turns out the only way in which things can go wrong is if $f'(f^{-1}(x))=0$. However, a lot more work is needed to show this than expected.
