Given the function $f: \mathbb{R}^3 \to \mathbb{R}$ defined by $f(x,\,y,\,z) := 3 x y^2 + 6 y^2 + x^2 z + 3 z^2 $ is not difficult to show that $f$ presents a unique critical point of coordinates $(0,\,0,\,0)$.
Unfortunately the Hessian matrix of $f$ in this point is positive semidefinite, then this point is relative minimum or "saddle".
Limiting restrictions $f(x,\,0,\,0) = 0, \; f(0,\,y,\,0) = 6y^2, \; f(0,\,0,\,z) = 3z^2$, I can state that this $(0,\,0,\,0)$ is a "saddle"'s point for $f$ because in the first restriction there is no critical point while in the second (and also in the third) in the origin there is a point of relative minimum?
Thanks!
Hint: The local minimum value, if it is one, would be $0$. Try setting $y = 0$ and then see what happens if you set $z = x^3$ for negative $x$ arbitrarily close to $0$.