The problem is the following:
Find a surface S that has a flat umbilical point P (this means $K = H = 0$ at P) such that for every open set $U \in \mathbb{R}^3, P\in U$, there are points of $S \cap U$ in each side of the tangent plane $P + T_P(S)$.
This problem is intented to be solved using geometric arguments, without the explicit need of proving that $S$ (once its found) has such property.
My professor has hinted at the use of a surface of revolution.
I, for starters, thought that if $K=H=0$ it meant that the surface is locally contained in a plane. Seems like I was wrong. Googling, I found the monkey saddle $z=x^{3}-3xy^{2}$ which verifies that $K=H=0$ at $(0,0,0)$ and $K<0$ at every other point, so this surface would work.
However I have no clue how should have I found this surface, and even if I had found it, it seems like it's not the logical path that my professor wanted me to make.
Any help on how to tackle this using a surface of revolution, or another method?
Let $f : \mathbb R^2 \to \mathbb R$ be a radially symmetric function so that $f(0,0) = f_i (0,0) = f_{ij} (0,0) =0$ (where $f_i, f_{ij}$ are the first and second partial derivatives of $f$). Then the graph
$$ \Gamma_f = \{ (x, y, f(x, y)) : (x, y)\in \mathbb R^2\}$$
is a surface of revolution in $\mathbb R^3$ and has a flat umbilical point at the origin.
Note that the tangent plane at the origin is the $x-y$ plane. So it suffices to find such $f$ so that $f$ changes sign infinitely many times at $(x, y) \to 0$. This can be found by (e.g.) $f(x, y) = g(\sqrt{x^2+y^2})$, where
$$ g(t) = \begin{cases} \sin\left( \frac{1}{t}\right) e^{-1/t^2}, & t\neq 0, \\ 0 & t=0.\end{cases}$$