I've got to check out if there is a holomorphic function $f$ such that $f(z)^3=z^3-1$ for all $a)$ $z \in B_1(0)$ and $b)$ $z\in B_1(1)$, where $B_r(z_0)$ is the open ball around $z_0$ with radius $r$.
I think the easiest way would be to take the Taylor series $f(z)= \sum_{n=0}^{\infty}a_nz^n$ and multiply out $f(z)^3$ in order to see if the coefficents there goes well with the cofficients of $z^3-1$. Of course it's a very exhausting way. Is there any more shorter way for this? Maybe working with logarithmus branch because of $f(z)=\sqrt[3]{z^3-1}$?
In $B_1(0)$, yes, there is such a function. If $z\in B_1(0)$, then $\lvert z\rvert<1$. Therefore, $\lvert z\rvert^3<1$ and so $z^3-1$ belongs to the halfplane $\{z\in\mathbb C\,|\,\operatorname{Re}z<0\}$. So, your idea is fine: you can work with an appropriate branch $\log$ of the logarithm and define $f(z)=\exp\left(\frac13\log\left(z^3-1\right)\right)$.
But in $B_1(1)$ there is no such function $f$. Note that $f(1)^3=1^3-1=0$. So, $1$ is a zero of $f$. Let $m$ be the order of that zero. But then the order of $1$ as a zero of $f^3$ is $3m$. This is impossible, since the order of $1$ as a zero of $z^3-1$ is $1$.