$f$ continuous, $g$ holomorphic with $g'(z) \neq 0 $, $g(f(z))=z$ $=> f$ holomorphic 
Let $U,V\subset \mathbb{C}$ be open, $f: U \to V$ continuous, $g\in H(V)$  with $g'(z)\neq0$ and $g(f(z))=z$. Then
  \begin{aligned} f \in H(U)\end{aligned}

My idea was to write:
$1=\lim_{z\to z_0} \frac{g(f(z))-g(f(z_0))}{z-z_0}=\lim_{z\to z_0} \frac{g(f(z))-g(f(z_0))}{f(z)-f(z_0)}*\frac{f(z)-f(z_0)}{z-z_0}$
Is it true, that since $\lim_{z\to z_0} \frac{g(f(z))-g(f(z_0))}{f(z)-f(z_0)}$ exists and is unequal $0$, also $\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ has to exist?
I think it does, but I am not sure and I cannot think of a proper way to proof this.
 A: Yes, you are right.
More rigorously, since the limit exist, we could write
$$\begin{align} 1 &= \lim_{z \to z_0} \frac{g(f(z))-g(f(z_0))}{f(z)-f(z_0)}\cdot \frac{f(z)-f(z_0)}{z-z_0}\\
&= \lim_{z \to z_0} \frac{g(f(z))-g(f(z_0))}{f(z)-f(z_0)}\cdot \lim_{z \to z_0}\frac{f(z)-f(z_0)}{z-z_0} \qquad (\star)\\
&= \lim_{w \to w_0} \frac{g(w)-g(w_0)}{w-w_0}\cdot \lim_{z \to z_0}\frac{f(z)-f(z_0)}{z-z_0},\end{align}$$
where $w_0 := f(z_0)$. 
Note that the second equality $(\star)$ is NOT the multiplication rule for limit! (Because we don't know if $\lim_{z \to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ exists or not.) Instead, the proposition we use is the following:

If $\lim_{z \to z_0} f_1(z)f_2(z)$ and $\lim_{z \to z_0} f_1(z)$ both exists and $\lim_{z \to z_0} f_1(z) \neq 0$,then $$\lim_{z \to z_0} f_1(z)f_2(z) = \lim_{z \to z_0} f_1(z) \cdot \lim_{z \to z_0} f_2(z)$$

The proof is not hard (by using $\epsilon-\delta$ argument.)
For the last equality, it holds since $f(z) \to f(z_0)$ as $z \to z_0$ by continuity of $f$.
Therefore $\lim_{z \to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ exists if $\lim_{w \to w_0} \frac{g(w)-g(w_0)}{w-w_0}$ exists and is unequal to $0$.
