Is there a holomorphic function such that $(f(z))^3=z-z^2$ Is there a holomorphic function $f:C-[0,1]$ such that $(f(z))^3=z-z^2$ for all $z\in C-[0,1]$

My intuition tells me that not really, for instance $$(z-z^2)^{\frac{1}{3}}$$

does not have a unique branch on this set, but I do not know how to formally prove it.
The function $z\to z-z^2$ has a double pole at infinity, which means that your $f$ would have a pole of order $2/3$ there!
The zeroes of $f$ must be the same as those of $z-z^2$, namely $0$ and $1$.

What is the winding number of $f(z)$ as $z$ describes a small circle around $0$? Since $z-z^2$ winds around $0$ once, you need an integer that gives $1$ when multiplied by $3$. But that is impossible.