How to check extrema if second derivative test fails I have to find minima and maxima of $f(x,y)=x^4+6y^2-4xy^3-1$
I found three points that could be extrema - $(0,0)$, $(1,1)$ and $(-1,-1)$
I already checked $(1,1)$ and $(-1,-1)$ with second derivative test, but determinant of Hesian at $(0,0)$ equals $0$ and I can't show that there is no extremum (but maybe there is?). What do I do now?

Hint: Assume that $x=\pm\epsilon$ and $y=\pm \delta$, where $\epsilon, \delta$ close to $0$ and assume without loss of generality that $\delta < \epsilon$. Then $$x^4+6y^2-4xy^3 \ge \epsilon^4+6\delta^2-4\epsilon\delta^3$$ Now for appropriate choice of $\epsilon$ and $\delta$ it is not difficult to see that the term $\epsilon^4+6\delta^2-4\epsilon\delta^3$ is always $>0$ close to $(0,0)$, since (for $\epsilon, \delta$ choosen appropriately) $$\epsilon^4+6\delta^2-4\epsilon\delta^3\ge\epsilon^4+6\epsilon^2-4\epsilon^4=3\epsilon^2(2-\epsilon^2)>0$$ so that in $(0,0)$ there is actually a minimum.
