It should be mentioned just how the critical point is determined, as it will be relevant to what follows. Setting the first partial derivatives of $ \ g(x,y) \ = \ x^3 \ + \ (x+y)^2 \ $ equal to zero yields
$$ g_x \ = \ 3x^2 \ + \ 2 \ (x+y) \cdot 1 \ = \ 0 \quad , \quad g_y \ = \ 2 \ (x+y) \cdot 1 \ = \ 0 $$
$$ \Rightarrow \ \ x \ + \ y \ = \ 0 \ \ \Rightarrow \ \ 3x^2 \ = \ 0 \ \ \Rightarrow \ \ x \ = \ 0 \ , \ y \ = \ 0 \ \ . $$
So $ \ (0, \ 0) \ $ is the only critical point on the surface $ \ z \ = \ g(x,y) \ $ , but there is something special about the line $ \ y \ = \ -x \ $ as well.
One way to evaluate the behavior of the function without graphing it is to take "cross-sections" on various planes containing the $ \ z-$ axis. In the $ \ yz-$ plane $ \ (x = 0) \ $ , our function becomes $ \ z \ = \ f(0, \ y) \ = \ y^2 \ $ , which we know is "concave upward" about the origin. However, in the plane $ \ y \ = \ -x \ $ , the "cross-sectional" function is $ \ z \ = \ f(x, \ -x) \ = \ x^3 \ $ , which has a change in concavity about the origin (point of inflection).
A point on a "surface" can only be a local extremum if the direction of concavity about that point is the same in every direction ("concave upward" in all directions for a local minimum and "downward" for a local maximum). This is not the case at $ \ (0, \ 0, \ 0) \ $ for our function, so it is a saddle point.
[Side note -- If we take other cross-sections through the surface on the planes $ \ y \ = \ 0 \ $ (the $ \ xz-$ plane) and $ \ y \ = \ x \ $ , we obtain the functions $ \ x^3 \ + \ x^2 \ $ and $ \ x^3 \ + \ 4x^2 \ $ , respectively. These have zero slope at the origin, but positive second derivatives. In these planes, $ \ (0, \ 0) \ $ is a local minimum, and we find that there is a nearby local maximum for small negative values of $ \ x \ $ . This appears to be exploring the developing "ridge" that appears toward the front of Amzoti's graph-cube. (Note that the "vertical" scale is highly compressed there.)]