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 $g$: 