Let $f : G \longrightarrow \Bbb C$ be a one-one and analytic function on a region $G$ and $f(G) = \Omega.$ Then $f^{-1} : \Omega \longrightarrow \Bbb C$ is also analytic and $$\left (f^{-1} \right )' (\omega) = \frac {1} {f'(z)},\ \text{where}\ \omega = f(z).$$
My attempt $:$ Since $f$ is one-one on an open set $G$ it has to be non-constant. So $f$ is a non-constant analytic function and hence by open mapping theorem it follows that $f$ is an open map. So $\Omega$ is an open set (since $\Omega = f(G)$) and also the map $f^{-1} : \Omega \longrightarrow \Bbb C$ is continuous on $\Omega.$ Also $f(f^{-1} (\omega)) = \omega,\ $ for all $\omega \in \Omega.$ Now fix some $b \in \Omega.$ Then there exists a unique $a \in G$ such that $f(a) = b.$ Since $\Omega$ is open $\exists$ $\delta \gt 0$ such that $B(b;\delta) \subseteq \Omega.$ Take any $h \in \Bbb C$ with $h \neq 0$ such that $|h| \lt \delta.$ Then $b + h \in B(b;\delta)$ and hence $b+h \in \Omega.$ So we have $f(f^{-1} (b)) = b$ and $f(f^{-1} (b+h)) = b+h.$ Since $h \neq 0$ it follows that $f^{-1} (b+h) \neq f^{-1} (b).$ Now we have \begin{align*} 1 & = \frac {f(f^{-1} (b+h)) - f(f^{-1} (b))} {h} \\ & =\frac {f(f^{-1} (b+h)) - f(f^{-1} (b))} {f^{-1} (b+h) - f^{-1} (b)} \frac {f^{-1} (b + h) - f^{-1} (b)} {h} \end{align*} Letting $h \to 0$ we find that $f^{-1} (b+h) \to f^{-1} (b)$ (since $f^{-1}$ is continuous on $\Omega$ and hence at $b \in \Omega$) and hence $\frac {f(f^{-1} (b+h)) - f(f^{-1} (b))} {f^{-1} (b+h) - f^{-1} (b)} \to f'(a)$ (since $f$ is analytic on $G$ and hence at $f^{-1} (b) = a \in G$). So if $f'(a) \neq 0$ then we find that $$\lim\limits_{h \to 0} \frac {f^{-1} (b+h) - f^{-1} (b)} {h} = \frac {1} {f'(a)},\ \text {where}\ b = f(a).$$ This implies that $(f^{-1})' (b) = \frac {1} {f'(a)},$ where $f(a) = b$ and $f'(a) \neq 0.$ So if we can show that $f'(z) \neq 0$ for all $z \in G$ then we are through. Why is it really the case?
Any help in this regard will be highly appreciated. Thanks in advance.
You can prove it in two steps: first show that zeros of $f'$ are discrete. Then use the Riemann removable singularity theorem to show that there are in fact none.
The zeros of $f'$ are discrete because otherwise $f'\equiv 0$ according to the identity theorem, which contradicts the fact that $f$ is injective. So the zeros of $f'$ are discrete.
Now since the zeros of $f'$ are discrete, so are their images under $f$: Let $f'(z_0)=0$. Because the zeros of $f'$ are discrete, there is an open neighborhood $U$ of $z_0$ which doesn't contain any other zeros of $f'$. Then due to the open mapping theorem $f(U)$ is an open neighborhood of $f(z_0)$, and due to injectivity of $f$, $U$ doesn't contain the image of any other zero of $f'$. So the images of the zeros of $f'$ are discrete.
You already showed that outside of these discrete points, $f^{-1}$ is analytic. And it is continuous even in those discrete points. The Riemann removable singularity theorem now states that $f^{-1}$ is in fact analytic in those points. You don't actually need to go further and prove that $f'(z)\neq0$ everywhere, since differentiability of $f^{-1}$ is what you actually wanted. But if you really want to: if $f^{-1}$ is differentiable, then the formula you derived holds. But that formula can only hold if the denominator is not $0$.