Obviously, if $f(z)$ is holomorphic on an open connected set $D$ and $f'(z)$ never vanishes, then around each point in $D$, we can find a neighbourhood in which $f(z)$ is invertible.
But what general conditions ensure that $f(z)$ is invertible on the whole of $D$? If, for instance, on the disc of radius $\frac{1}{2}$ around $(1+0i)$, we cannot invert $f(z)=z^{100}$.
It will suffice, I suppose, if $\Re (f'(z))$ never attains $0$. But no stronger conditions come to mind.
The inequality $\operatorname{Re}f'>0$ is sufficient for global invertibility in a convex domain, where one can integrate it along line segments to obtain $\operatorname{Re} \frac{f(a)-f(b)}{a-b}>0$ for all $a\ne b$. In general, it is not sufficient when the domain is not convex.
There are many sufficient univalence criteria (search for this term) for holomorphic functions on the unit disk. They often involve the Schwarzian derivative $Sf(z) = (f''/f')' -\frac12(f''/f')^2$. The most notable is $|Sf(z)| \le 2/(1-|z|^2)^2$; Nehari's paper Univalence criteria depending on the Schwarzian derivative gives a few others.