This is exercise 9 of chapter III.4 in Freitag and Busam.
$\mathbb{E}:=\{z\in\mathbb{C} | |z|<1\}$. Let $f,g:\mathbb{E}\rightarrow \mathbb{E}$ be bijective analytic functions, which satisfy $f(0)=g(0)$ and $f'(0)=g'(0)$. Moreover, assume that $f'$ and $g'$ have no common zero. Show that $f(z)=g(z)$ for all $z\in\mathbb{E}$.
The indication was to use Lemma III.3.8 for the function $f\circ g^{-1}$. The lemma states: Let $\varphi:\mathbb{E}\rightarrow \mathbb{E}$ be a bijective map with zero as a fixed point, such that both $\varphi$ and $\varphi^{-1}$ are analytic. Then there exists a complex number $\zeta$ with modulus $1$ such that $$\varphi(z)=\zeta z.$$
My question is, how do we know if $f\circ g^{-1}$ is analytic and has an analytic inverse? Is $f^{-1}$ analytic? Why do we need to assume that $f'$ and $g'$ have no common zero?
Use this answer to see why $f^{-1}$ should be analytic (and in the way, why $f'$ and $g'$ never vanish anyway).
Then, after applying the lemma to $f^{-1}\circ g$, you have $g'(0)=\zeta f'(0)$, this is where you need $f'(0)$ and $g'(0)$ to not be zero, so that $\zeta =1$.