The function $w:=f(z)$ defined implicitly by $\Phi(w,z)=w^2-z^2-z^3=0 $ has two critical points, $z=0$ and $z=-1$. I thought both of them were branch points (and hence singularities) but I realized it's possible to expand (Taylor series) both branches about $z=0$:
$f_1(z)=\sqrt{z^2+z^3}=z+\frac{1}{2}z^2-\frac{1}{8}z^3$+...
$f_2(z)=-\sqrt{z^2+z^3}=-f_1(z)$
Thus is $f(z)$ analytic at $z=0$? If not, how is it possible to have a Taylor series around a singularity?
Yes, it's analytic at $z=0$. You can think of this as the product of $\sqrt{z^2}$ and $\sqrt{z+1}$.
The $\sqrt{z^2}$ part clearly has 2 branches ($\pm z$) which are analytic everywhere.
For the $\sqrt{z+1}$ part, you can choose a branch which is analytic at $z=0$ (there would only be a problem if you were centered at $z=-1$).