Find local extrema of $f(x,y)=y^2-3x^3y+2x^6$
My attempt:
First we equal the partial derivatives to zero to find the estacionary points:
$D_1f(x,y)=0 \rightarrow -6yx^2+12x^5=0$
$D_2f(x,y)=0 \rightarrow 2y-3x^3=0$
The only estacionary point is $(x, y)=(0,0)$
Then we calculate the Hessian matrix:
$D_{11}f(x,y)=-12yx+60x^4$
$D_{22}f(x,y)= 2$
$D_{12}f(x,y)=D_{21}f(x,y)=-6x^2$
Then $Hf(0,0)=\begin{bmatrix}D_{11}f(0,0) & D_{12}f(0,0)\\D_{21}f(0,0) & D_{22}f(0,0)\end{bmatrix}=\begin{bmatrix}0 & 0\\0 & 2\end{bmatrix}$, so $det(Hf(0,0))=0$.
Because $det(Hf(0,0))=0$ we cannot determine whether $(0,0)$ is an extremum or not by studying its Hessian matrix.
After this I've tried to show $(0,0)$ is not an extremum by showing there are points greater than $f(0,0)$ and smaller than $f(0,0)$ in every neighborhood of $(0,0)$.
If $x>0, y=0$ then $f(x,y)=2x^6>f(0,0)=0$ So there are points $(x,y)$ arbitrarily close to $(0,0)$ where $f(x,y)>0=f(0,0)$.
I have failed to find points near $(0,0)$ whose image is smaller than $f(0,0)$ near to $(0,0)$
For $y=0$ we have $$f(x,y)=2x^6$$so $(x,y)=(0,0)$ is a minimum on this branch of function. Also: $$f(x,y)=y^2-3x^3y+2x^6=(y-1.5x^3)^2-0.25x^6$$so on branch $y=1.5x^3$ we have $f(x,y)=-0.25x^6$ and $(x,y)=(0,0)$ is a maximum on this branch so is a saddle point.