Consider the function $$f(x, y) = x^4+y^4-y^2$$
After having compute the gradient, I found the following critical points
$$p_0 = (0, 0) \qquad \qquad p_{1, 2} = \left(0, \pm \sqrt{\frac{1}{2}}\right)$$
Now I followed this question: Characterizing the critical point of a two-variable function when the Hessian determinant is zero to understand how to work with $(0, 0)$ and I got it.
The other points though, I need confirmation.
The function is even, so the two points must be the same type.
Now, I thought of this: the hessian reads
$$H = \begin{pmatrix} 12 x^2 & 0 \\ 0 & 12y^2 - 2 \end{pmatrix}$$
I could study the determinant and find where it's positive, hence where the function is convex.
$$12x^2(12y^2-2) > 0$$
Which gives the solution $\forall x\in\mathbb{R}$ and $y > \sqrt{\frac{1}{6}}$ and $y < -\sqrt{\frac{1}{6}}$
Now, since my critical points are beyond the solutions, I can say that the determinant of the Hessian is positive at $\left(1, \pm \sqrt{\frac{1}{2}}\right)$, (for every neighbourhood $\epsilon$ of zero of $x$) hence the two points are two minima.
I wanted to understand also the other way (assuming this is correct), that is the restrictions as done in the linked question. But here I got stuck, because my function would read:
$$f\left(\epsilon, +\sqrt{\frac{1}{2}}\right) = \epsilon^4 - \frac{1}{4}$$
Which is always negative for every neighbourhood $\epsilon \to 0$.
Where am I wrong?