Is $(0,0,0)$ saddle point? $f(x,y,z)=x^3+y^3+z^3-3xy-3yz$
Is $(0,0,0)$ saddle point?
$\nabla f(0,0,0)=0$, so $(0,0,0)$ is one of the stationary point.
Also, because of my posture The reason why $f(0,0,0)$ is not a extreme value. ,
$f(0,0,0)$ is not a extreme value.
Therefore, I think $(0,0,0)$ is a saddle point.
However, according to wolfram alpha, saddle points of $f(x,y,z)=x^3+y^3+z^3-3xy-3yz$ don't exist.

The Hessian at $(0,0,0)$ is:
\begin{pmatrix}
0 & -3 & 0 \\
-3 & 0 &-3 \\
0 & -3 & 0
\end{pmatrix}
Its npn-zero eigenvalues are $\pm3\sqrt2$. So by definition it is indefinite (not positive or negative definite) and so $(0,0,0)$ is a saddle point.
