I'm wondering if $\sqrt[3]{z^2}$ is a Riemann surface including point $(0, 0)$ or not.
Treat it as the zero set of $F(z, w) = 0$, where $F(z, w) = z^2-w^3$. Then $F_z = 2z$ and $F_w = 3w^2$ and both are zero in point $(0, 0)$. From this view point it seems it doesn't define a Riemann surface since I cannot apply the implicit function theorem. Hence the Riemann surface it defined doesn't include the point $(0, 0)$.
But if I follow another viewpoint, treating it as a complete analytic function, it seems that it's a Riemann surface and can be extended to include the point $(0,0)$ as a branch point. And it seems similar with the riemann surface defined by $\sqrt[3]{z}$, which has the point $(0, 0)$ as a branched point.
Is that I missed something? Does the riemann surface including point $(0, 0)$ or not?
Thanks!