Sign of $K$ does not imply local convexity

We know that if $K(p)>0$ at one pint, then the surface locally in one side of its affine tangent plane. On the other hand, if the surface lies in one side around a point, then $K(p)\geq 0$. However, there are surfaces that at one point $p\in S$, $K(p)=0$, $K>0$ around $p$, BUT the surface lies in both sides of $T_pS$. An example is the surface $z=f(x,y)=x^3(1+y^2)$. Let $p=(0,0,0)$. Take the parametrization $$X(x,y)=(x,y,x^3(1+y^2)).$$ Then $X^{-1}(p)=(0,0)$ and as $X_x(0,0)=(1,0,0)$ and $X_y(0,0)=(0,1,0)$, then $T_pS$ is the plane $z=0$. Since $1+y^2\geq 0$ and $x^3$ changes of sign at $x=0$, then $f$ change of sign around $p$, that is, the surface has points in both sides of $T_pS$, as it appears in the next figure:

We now compute the Gauss curvature using the formula $$K(X(x,y))=\frac{f_{xx}f_{yy}-f_{xy}^2}{(1+f_x^2+f_y^2)^2}.$$ Then $$K(X(x,y))=\frac{12x^4(1-2y^2)}{(1+f_x^2+f_y^2)^2}.$$ In the open set of $S$ gieven by $V=X(U)$, where $U=\{(x,y): x\in{\mathbb R},|y|<1/\}$, $K>0$ in $V-\{p\}$ and $K(p)=0$. The next figure is the numerator of $K$ in $|x|<1$, $|y|<1/2$ that hows that the sign of $K$ is, indeed, positive.