In dealing with your first question about the specific function, the computational approach is to find the critical point(s) from
$$ f_x \ = \ 2x \ = \ 0 \ \ , \ \ f_y \ = \ 2y \ = \ 0 \ \ , \ \ f_z \ = \ -2z \ = \ 0 \ \ , $$
giving us the origin $ \ (0, \ 0, \ 0) \ $ as the only critical point. The non-zero second partial derivatives are $ f_{xx} \ = \ 2 \ , \ \ f_{yy} \ = \ 2 \ , \ \ f_{zz} \ = \ -2 \ $ , produce a diagonal Hessian matrix
$$ \left[ \begin{array}{ccc}2&0& \ 0\\0&2& \ 0\\0&0&-2\end{array} \right] \ \ , $$
from which we can simply "read off" the eigenvalues as $ \ 2, \ 2, \ $ and $ \ -2 \ $ . Since these are not all positive or all negative, the critical point is a saddle point.
As is often the case with functions that have a high degree of symmetry, we can find a geometrical interpretation that make this reasonably easy to see. In his comment above, Martin Sleziak suggests something along this line.
The function $ \ f(x, \ y, \ z) \ = \ x^2 \ + \ y^2 \ - \ z^2 \ $ is defined everywhere in $ \ \mathbb{R}^3 \ $ , so we will consider its "level surfaces". The "zero surface" $ \ x^2 \ + \ y^2 \ - \ z^2 \ = \ 0 \ $ is the two nappes of the cone $ \ x^2 \ + \ y^2 \ = \ z^2 \ $ . Anywhere "inside" either of these nappes, $ \ |z| \ > \ \sqrt{x^2 + y^2} \ $ , which is the perpendicular distance of a point from the $ \ z-$ axis ; so $ \ f(x, \ y, \ z) \ < \ 0 \ $ within the nappes, and tends to negative infinity as we "move" farther from the origin in that region. Outside of the nappes, $ \ |z| \ < \ \sqrt{x^2 + y^2} \ $ , thus $ \ f(x, \ y, \ z) \ > \ 0 \ $ and increases without limit as we "go farther away" from the origin in that region.
We can imagine then any sort of "path" from a distant point outside the nappes to the origin, and thence to a distant point inside them; we could even "travel" along the surface of the cone for some part of that part. So there is nothing special whatever about the "critical point" at the origin; our function thus has neither global nor relative extrema.