Show that $f$ has no local extremum in $(0,0)^T$

Question

I have the following problem.

Let $f : ℝ^2 → ℝ$ be given by $f(x,y) = x^2 - 3xy^2 +y^4.$ Explain why $f$ in $(0,0)^T$ has no local extremum.

Attempt:

So unfortunately I just come not on that. I already calculated the partial derivatives and also set up the Hessian matrix for this by setting $f_x$ and $f_y = 0$. For this I only got $x = 0$ and $y = 0$ as a point $(0,0)$, and

$f_x = 2x-3y^2,$
$f_y = 4y^3-6xy,$
$f_{xx} = 2,$
$f_{yy} = 12y^2-6x,$
$f_{xy} = f_{xy} = -6y.$

So for the matrix at the point $(0,0)$ I then have the matrix $\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}$ and for their eigenvalues ​​then $λ_1 = 0$ and $λ_2 = 2$ and I think that's semi positive definite and now i dont know how to show, theres no local extremum.

Answer 1

If $(x,y)\in\Bbb R^2$, then $x^2-3xy^2+y^2=(x-y^2)^2-xy^2$. Therefore, $f(y^2,y)=-y^4<0$ (unless $y=0$). But $f(x,0)=x^2>0$ (unless $x=0$).

Answer 2

Take the direction $x=y^2$. $f(y^2, y)=-y^4<0$ in an open neighborhood of $0$. Also $f(x, 0)=x^2>0$ near zero. So its not a local extremum