Find a branch $g(w)$ of $w^{1/3}$ which serves as the inverse function $f(z)=z^3$ with $\text{dom}(f)=\{z\in\mathbb{C}^\times\mid \text{Arg}(z)\in[0,2\pi/3)\}$. Relate $g(w)$ to the principle branch $\sqrt[3]w$.
So I've identified that I should take the domain to be $\mathbb{C}^+=\mathbb{C}\setminus [0,\infty)$, and then do something like $$g(w)=\zeta \sqrt[3]{|w|}e^{i\text{Arg(w)/3}}=\frac{-1+i\sqrt{3}}{2}\sqrt[3]{|w|}e^{i\text{Arg(w)/3}}.$$ But this is just a guess based on the same kind of thing for $w^{1/2}$ so it's is probably wrong.
Cut the $w$-plane from $0$ to $\infty$ along the real axis. Note that $w^{1/3}$ is analytic on $\mathbb{C}\setminus [0,\infty)$. Choose the branch for which $0\le \arg(w)<2\pi$.
Let $w=z^3$. Then $0\le \arg(z)<2\pi/3$.
Finally, $\arg(w)=\text{Arg}(w)$ in the upper-half plane and $\arg(w)=\text{Arg(w)}+2\pi$ in the lower-half plane. So, we have
$$g(w)=\begin{cases} |w|^{1/3}e^{i\text{Arg}(w)/3}&,0\le\text{Arg}(w)<\pi\\\\|w|^{1/3}e^{i2\pi/3}e^{i\text{Arg}(w)/3}&,-\pi\le\text{Arg}(w)<0\end{cases} $$
And we are done.