I am trying to find the correct branch of the complex cube root, and map the complex plane cut along the positive y axis onto the wedge $-\frac{\pi}{2}<\arg(w)<\frac{\pi}{6}$. The function $w=\sqrt[3]{z}$ clearly does the job, however one has to choose the proper branch. I am doing this by guessing.
Choose
$$z=-i=e^{i\left(\frac{3\pi}{2}+2\pi k\right)},$$
which is in the original plane. Therefore, $k=-1$ does the job, since the argument of $\sqrt[3]{z}$ with the so chosen branch $$ \frac{1}{3}\left(\frac{3\pi}{2}-2\pi\right)=-\frac{\pi}{6}, $$ is in the range $-\frac{\pi}{2}<\arg(w)<\frac{\pi}{6}$. In general,
$$f(z)=e^{i\frac{1}{3}\left(Arg(z) - 2\pi\right)},$$
maps the the complex plane except the positive y axis to the angular segment $-\frac{\pi}{2}<\arg(w)<\frac{\pi}{6}$. Here $Arg(z)$ is the principal value of the argument $Arg(z) \in (-\pi,\pi]$ or $Arg(z) \in [0,2\pi)$
In general, the solution goes as follows. The complex cube root has 3 values $\frac{\pi}{6}$ for $k=0$,$\frac{5\pi}{6}$ for $k=1$ and $\frac{3\pi}{2}$ for $k=2$. That is, the cube root divides the complex plane into 3 equal angular segments.
Only the value $k=0$ touches the range of interest $-\frac{\pi}{2}<\arg(w)<\frac{\pi}{6}$ , therefore, one has to go one revolution back and choose $k=-1$.
Why $k=-1$ is the part I do not understand.
You should be aware that $\arg$ has infinitely many branches and thus $a <\arg(w)< b$ needs interpretation. You will certainly not like to have the branch cut of $\arg$ in the above region. Thus I suggest to consider an open interval $(a,b)$ such that $b - a \le 2\pi$ and define $$G(a,b) = \{ re^{it} \mid r > 0, t \in (a,b) \}$$ instead of $G(a,b) = \{ z \mid a < \arg(z) < b \}$. Obviously we have $G(a,b) = G(a +2k\pi,b+2k\pi)$ for all $k \in \mathbb Z$. Note that $$\mathbb C(c) = G(c,c+2\pi) = \mathbb C \mathbb \setminus L(c)$$ with $L(c) = \{ re^{ic} \mid r \ge 0 \}$ is a sliced plane with branch cut at an angle of $c$.
On each $\mathbb C(c)$ we have three distinct holomorphic branches $\beta^c_k$ of $\sqrt[3]{z}$ given by $\beta^c_k(re^{it}) =\sqrt[3]{r}e^{i(t + 2k\pi)/3}$ for $k = 0,1,2$ and $t \in (c,c+2\pi)$. Their range is $G^3_k(c) = G(c/3+k\pi/3,c/3+(k+1)\pi)$. Note that replacing $c$ by $c + 2\pi$ does not change the set $\mathbb C(c)$ and we get the same collection of three cubic root maps on $\mathbb C(c)$, but we have a cyclic index shift ($\beta^{c+2\pi}_k = \beta_{k+1}^c$, where we understand $2+1 = 0$).
Let us now consider the region $G(a, a + 2\pi/3)$. It is mapped by $z^3$ onto $\mathbb C(3a)$. Both sets only depend on $a \mod 2\pi$. The branch $\beta_0^{3a}$ maps $\mathbb C(3a)$ onto $G(a, a + 2\pi/3)$.
In your case we have $a = -\pi/2$, thus the desired branch cut is at $-3\pi/2$ which is the same as at $\pi/2$. Thus the sliced plane is $\mathbb C \setminus$ positive $y$-axis. Note, however, that for the branch cut at $-3\pi/2$ we have to take $$\sqrt[3]{z} = \beta_0^{-3\pi/2}(z), \text{ i.e. } \sqrt[3]{re^{it}} = \sqrt[3]{r}e^{it/3} \text{ for } t \in (-3\pi/2,\pi/2)$$ and for the branch cut at $\pi/2$ we have to take $$\sqrt[3]{z} = \beta_2^{\pi/2}(z), \text{ i.e. } \sqrt[3]{re^{it}} = \sqrt[3]{r}e^{i(t+4\pi)/3} \text{ for } t \in (\pi/2,5\pi/2)$$ which is the same although it looks different.