Surface with only one parabolic point

We have proved that at an elliptic point, the surface lies in one side of the tangent plane at that point. Exactly, if $K(p)>0$, then there exists a neighborhood $V\subset S$ of $p$ such that $V\cap T_pS=\{p\}$ and $V-\{p\}$ lies in one of the two open halfspaces determined by $T_pS$. 

Here we show a surface with the same property but $K(p)=0$. The surface is obtained by rotating the curve $z=x^4$ around the $z$-axis. At the point $p=(0,0,0)$, $K(p)=0$, the tangent plane $T_pS$ is the $xy$-plane and $S-\{p\}$ lies in the halfspace $z>0$. Exactly, with the usual parametrization $X(x,s)=(x\cos(s), x\sin(s), x^4)$, we have 

$$K(x,s)=\frac{36 x^4}{(1+16x^6)^2},$$

so $K(p)=0$ and the rest of points are elliptic, that is, $p$ is the only parabolic point of $S$.