How to caculate the inverse of function $x^3$ in $\mathbb{F}_{2^n}$

The order of the multiplicative group of $\mathbb F_{2^n}$ is $2^n-1$. If 3 is coprime to $2^n-1$ then there exists $d\in [1,\ldots,2^n=1]$ such that $3d\equiv 1\pmod{2^n-1}$. We can find such a $d$ using the extended Euclidean algorithm.

The function on $\mathbb F_{2^n}$ $y\mapsto y^d$ is then the inverse of the map $x\mapsto x^3$ since for $x\in\mathbb F_{2^n}^\times$ we have $(x^3)^d=x^{3d}=x^1$ (the case $x=0$ is obvious).

This means that $x\mapsto x^3$ is bijective and hence a permutation.

In the case where $3|2^n-1$, if $x^3=y$ in $\mathbb F_{2^n}$ then so to does $(\omega x)^3=y$ and $(\omega^2 x)^3=y$ where $\omega$ is a cube root of 1 in $\mathbb F_{2^n}$. It follows that in this case the map is not injective and so not a permutation.