I am trying to find the critical points of
$$f(x,y)=(x^2-y^4)(1-x^2-y^4).$$
There are 9, and I found all their coordinates. I’ve classified 6/9 using the hessian, but the remaining 3 have at least one eigenvalue of 0. I can’t find anything that explains what I should do in this situation. The markscheme to the question states ‘need to see $y$ variation is $y^4$ which is positive’ but I don’t understand what that means.
I would love some help getting a little more of an intuitive grasp as to what is going on, as well as how on Earth I’m supposed to solve it.
$(0,0)$ is a stationary point for which the Hessian is $\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}$, with eigenvalues $2$ and $0$. Now note that $$f(x,0) = x^2(1-x^2).$$ So increasing $x$ slightly from $(0,0)$ causes an increase in $f$. Also, $$f(0,y) = -y^4(1-y^4),$$ so increasing $y$ slightly from $(0,0)$ causes a decrease in $f$. Thus $(0,0)$ must be a saddle point.
At $(1/\sqrt 2, 0)$, the Hessian is $\begin{pmatrix} -6 & 0 \\ 0 & 0 \end{pmatrix}$, with eigenvalues $-6$ and $0$. One can show that, with a Taylor expansion, $$f(x, y) = \frac 14 - 2\left(x - \tfrac{1}{\sqrt 2}\right)^2 - 2 \sqrt 2\left(x - \tfrac{1}{\sqrt 2}\right)^3 - \left(x - \tfrac{1}{\sqrt 2}\right)^4 - 4 y^4 + O\left(\left(x - \tfrac{1}{\sqrt 2},y\right)^5\right).$$ Thus $(1/\sqrt 2, 0)$ is a local maximum, and by symmetry so is $(-1/\sqrt 2, 0)$.