Q:

Classifying the stationary point of $f(x, y) = 2x^4-3x^2y+y^2$

Find and classify the stationary point on the surface $f(x, y) = 2x^4-3x^2y+y^2$

I found the only stationary point for this surface is at $(0,0)$ but the determinant of the Hessian matrix for this surface is zero and I don't know how to proceed with this problem.

A:

Indeed $(0,0)$ is the only critical point of $f$. Since the Hessian at $(0,0)$ is singular we need algebraic tricks in order to characterize the behavior of $f$ at this point. In the case at hand the trick consists in writing $f$ in the form
$$f(x,y)=(2x^2-y)(x^2-y)\ .$$
This shows that $f(x,y)\equiv0$ along two parabolas touching at $(0,0)$. These parabolas partition the plane into four subsets in which $f(x,y)$ has eiter positive or negative sign. The point $(0,0)$ belongs to the boundary of each of these subsets. It follows that you can call $(0,0)$ a saddle point of $f$. In any case $(0,0)$ is neither a local minimum nor a local maximum of $f$.

