Let $$f(z) = C \int_{0}^{z} \frac{1}{(1-\zeta^3)^{\frac{2}{3}}}d\zeta$$ with $$C \int_{0}^{1} \frac{1}{(1-x^3)^{\frac{2}{3}}}dx=1$$ be a conformal map from the unit disc $B(0,1)$ to the equilateral triangle with the corners $e^{\frac{2k\pi i}{3}}, k = 0,1,2$. Show that there exists a meromorphic function $g: \mathbb C \to \mathbb C$, for which $f(g(z))=z$, first for $z$ from the triangle and then, by using analytic continuation, for all $z$.
Let $\mathbb T$ denote the set of all $z$ that belong to the triangle.
I know that a meromorphic function is, in this case, a holomorphic function on $\mathbb T$ except for the pole points, i.e. the triangle corners.
A first naive try would be to determine $f^{-1}(z)$, since $f^{-1}(f(g(z))) = g(z)$. But I'm not sure how to get $f^{-1}(z)$, if this is even the right way to do it. Any help would be appreciated.