Proving the origin is a saddle point. I have the function $g(x,y) = x^6 -y^6x^2$ and want to prove that the origin is a saddle point.
I know that a critical point with an indefinite Hessian matrix is a saddle point, but this is only a sufficient condition.
$(0,0)$ is indeed a critical point of $g$, but the Hessian matrix is everywhere $0$ and hence it is positive and negative semi-definite and so not indefinite. 
How would I go about concluding that the origin is indeed a saddle point here? Unfortunately, the only definitions of saddle points that I could find gave the usual sufficient condition of an indefinite Hessian.
Many thanks.
 A: On the line $y=0$, $g(x,y)=x^6$, which is concave up.
On the curve $x=y^2$, $g(x,y)=y^{12}-y^{10}=y^{10}(y^2-1)$, which is concave down.
More details, as requested:
A saddle point is stationary, but neither a local max nor a local min.  $g(x,y)$ is stationary at the origin, because both partials are zero.  $(0,0)$ is not a local max by the first observation above, it is not a local min by the second one.
A: We have
$$g(x,y)=x^2(x^4-y^6)=|x|^2\bigl(|x|^4-|y|^6\bigr)\ ,$$
in particular $g(0,0)=0$. Furthermore $g(x,y)<0$ when $0<|x|<|y|^{3/2}$ and $g(x,y)>0$ when $|x|>|y|^{3/2}$. There are both kinds of such  $(x,y)$ near $(0,0)$; therefore $g$ has neither a local maximum nor a local minimum at $(0,0)$. Whether you want to call $(0,0)$ a saddle point is up to you; at any rate such a saddle would not be very comfortable to sit on: Going in a small circle around $(0,0)$ you will experience four maxima and four minima.
The point $(0,0)$ is a degenerate critical point of $g$. Such points ${\bf p}$ are characterized by $\nabla f({\bf p})=0$, $\det H({\bf p})=0$. Their analysis requires special treatment, which was easy in the case at hand.
