I've read in a book that the surface of revolution of the curve $y=z^4$ around the $z$ axis is an example of a regular surface which has a plane point (in the origin), but it is completely contained in one side of the tangent plane in that point. This shows that the reciprocal of a well known result about elliptic and parabollic points is not true.
Now, how can I show that this surface of revolution is, in fact, a regular surface?
The standard parametrization is not one to one (because the curve intersects the axis of rotation) and I cannot find another one that works, nor another method to prove it is a regular surface.
Thanks a lot for the answers.
EDITED: I think I've found an atlas for the surface. $$X_1(u,v)=\begin{cases} (v \cos(u),v\sin(u),v^4) & \text{if} & (u,v) \in \left((0, \pi) \setminus\{\frac{\pi}{2}\} \right) \times \left( \Bbb R \setminus \{0\} \right)\\ (0,0,0) & \text{if} & (u,v)=\left(\frac{\pi}{2},0\right) \end{cases}$$
$$X_2(u,v)=\begin{cases} (v \cos(u),v\sin(u),v^4) & \text{if} & (u,v) \in \left((\frac{\pi}{3}, \frac{4\pi}{3}) \setminus\{\frac{5\pi}{6}\} \right) \times \left( \Bbb R \setminus \{0\} \right)\\ (0,0,0) & \text{if} & (u,v)=\left(\frac{5\pi}{6},0\right) \end{cases}$$
They are not regular parametrizations, their derivatives on $u$ are null in $\left(\frac{\pi}{2},0\right)$. So, I'm again at the starting point.
Alternatively, note that your surface can also be expressed as the graph of the function $$z(x,y) = (x^2+y^2)^2$$ over the whole plane $\Bbb R^2$ (why?), and graphs are automatically regular surfaces: we have a global parametrization $(x,y)\mapsto (x,y,z(x,y))$ with its inverse given by projection, $(x,y,z(x,y))\mapsto (x,y)$.