$$f(x,y)=x^4+y^4-2x^2+4xy-2y^2$$
I have found out that $(\sqrt{2},-\sqrt{2})$ and $(\sqrt{2},-\sqrt{2})$ are minimum point, but for $(0,0)$ I get that the Hessian is
$$\begin{pmatrix} -4 & 4 \\ 4 & -4 \end{pmatrix}$$
Which has determinant of zero, what should I do to determine which point is it?
We have $f(x,x)=2x^4$ and $f(x,-x)=2x^2(x^2-4)$. Hence, for $x \ne 0$ in a "small" neighborhood of $0$ we have
$f(x,-x) <0 =f(0,0) < f(x,x)$.
Thus $(0,0)$ is a saddle point.