Critical Points. Find and classify. Given $g(x,y)=y^2 - x^3$
find the critical points and classify them
$$\nabla g(x,y) = \begin{pmatrix}
        -3x^2 \\
        2y \\
        \end{pmatrix}$$
So, 
$\implies  -3x^2=0,2y=0$
$\implies  x=0,y=0$
$$Hf(x,y) = \begin{pmatrix}
        -6x & 0 \\
        0 & 2 \\
        \end{pmatrix}$$
$$Hf(0,0) =\begin{pmatrix}
        0 & 0 \\
        0 & 2 \\
        \end{pmatrix}$$
$det( Hf(0,0) ) = 0$
Is this correct? 
How would I finish this and interpret the results? 
For example, whether or not it's positive-semi definite.
Thanks
 A: A critical point $(x,y)$ for $g$ is such that the vector $(\nabla g)(x,y)$ equals to $\vec{0}$. That is, both coordinates must be equal to zero. In your case, you must have both $-3x^2 = 0$ and $2y = 0$. This means that the only critical point is $(0,0)$. Plugging it into the Hessian, and taking the determinant you obtain $0$. So the determinant doesn't provide you enough information on the type of the critical point.
Notice that the value of $g$ at the critical point is $0$. What you can do to determine the type of the critical point is to plot the contour $g(x,y) = 0$ on which the critical point lies. It looks like:

On the contour, the function $g$ is $0$. What is the value of $g(x,y)$ on a point $(x,y)$ that lies to the right of the contour? You can take any point, and just plug it into $g$ to check the sign of $g$. Say, $g(1,0) = -1$, so on the right of the contour the function $g$ is negative. Similary, checking $g(-1,0) = 1$, we see that on the right of the contour the function is positive. This means that the critical point is a saddle point. This is confirmed by the graph of $g$:

