Show that $(0,0)$ is not a local minimum of $f(x,y)=(y-x^2)(y-2x^2)$

Question

Show, that $f(x,y)=(y-x^2)(y-2x^2)$ reaches it's minimum at $M(0,0)$ for $f(t\cos\alpha,t\sin\alpha)$, but $M$ is NOT a local minimum point.

Seemed to be an easy one, but I couldn't handle it.
We have.

$f(t)=(t\sin\alpha-t^2\cos^2\alpha)(t\sin\alpha-2t^2\cos^2\alpha)$ for any fixed $\alpha$

$f'(t)=8t^3\cos^4\alpha-9t^2\sin\alpha\cos^2\alpha+2t\sin^2\alpha=0$

$t_0=0$ is a root for $f'$, and $f'(0)=0$, however it has 2 more roots since $D>0$

$f''(t_0)=24t_0^2\cos^4\alpha-18t_0\sin\alpha\cos^2\alpha+2sin^2\alpha=2sin^2\alpha>0$
so $t_0$ is a minimum. part 1 done.

$$\begin{cases} \frac{\partial f}{\partial x}=2x(4x^2-3y)=0 \\ \frac{\partial f}{\partial y}=2y-3x^2=0 \end{cases} $$ solution is again $M(0,0)$

$$\frac{\partial^2 f(M)}{\partial x^2}=24x^2-6y=0,\frac{\partial^2 f(M)}{\partial y^2}=2,\frac{\partial^2 f(M)}{\partial x \partial y}=-6x=0$$

$$\Delta = \begin{vmatrix} \frac{\partial^2 f(M)}{\partial x^2} & \frac{\partial^2 f(M)}{\partial x \partial y} \\ \frac{\partial^2 f(M)}{\partial x \partial y} & \frac{\partial^2 f(M)}{\partial y^2} \end{vmatrix}=0$$ I've got the determinant is zero for the particular point, and here I reached to dead end, I have no idea how to continue/conclude, please help me to finish this one. thanks

Answer 1

Every neighbourhood of the origin contains points $(x,y)$ for which $f(x,y)>0$ and other points $(s,t)$ for which $f(s,t)<0$, so $(0,0)$ cannot be a local minimum.

Answer 2

Approach the origin along the curve $y=\frac{3}{2}x^2$. For points along this curve the origin is a local maximum, so the origin can't be a local minimum.