Finding the critical points of $f(x,y)=(y-x^2)(y-2x^2)$

Finding the critical points of $f(x,y)=(y-x^2)(y-2x^2)$

I know that $(a,b)$ is a critical point $\iff \nabla f(a,b)=(0,0)$

So $\nabla f(x,y)= (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$

$\nabla f(x,y)=(8x^3-6xy, -3x^2+2y)=(0,0) \iff (x,y)=(0,0)$

So finding the Hessian matrix at $(0,0)$

$\frac{\partial ^2f}{\partial x^2}[(y-x^2)(y-2x^2)]=\frac{\partial
f}{\partial x}[8x^3-6xy]=24x^2-6y$
$\frac{\partial ^2f}{\partial y^2}[(y-x^2)(y-2x^2)]=\frac{\partial f}{\partial y}[-3x^2+2y]=2$
$\frac{\partial ^2f}{\partial x \partial y}[(y-x^2)(y-2x^2)]=\frac{\partial
f}{\partial x}[8x^3-6xy]=-6x$

$H(0,0)=\begin{pmatrix}
0 & 0 \\
0 & 2
\end{pmatrix}$

Finding the eigenvalues for $H(0,0)$:

$\det(H(0,0)-\lambda I)= \begin{vmatrix}
-\lambda & 0 \\
0 & 2-\lambda
\end{vmatrix} = \lambda^2-2\lambda$

$\lambda^2-2\lambda=0 \iff \begin{cases}
\lambda_1 = 0 \\
\lambda_2 = 2
\end{cases}$

So I can't really tell if $f(x,y)$ has any relative maximum nor minimum or a saddle point, and I don't know how to continue…

Edit: I know that this is a duplicate post, but I don't understand the answer, and the question is 4 years old.