Answer
$(-1,-1)$ is a local minimum.
$(-1,0)$ is a saddle point.
$(-1,1)$ is a local minimum.
$(1,-1)$ is a saddle point.
$(1,0)$ is a local maximum.
$(1,1)$ is a saddle point.
Work Step by Step
It looks as though the graph has six critical points. It seems as though it has a local extrema at $(-1,1)$, $(-1,-1)$, and $(1,0)$. It seems as though the points where $x=-1$ are local minimum points because the value of the level curve decreases as we get closer to the points. The one point with $x=1$ looks like a local maximum point because the value of level curve increases as we get closer to the point.
Next, it looks as though it has a saddle point at the points, $(-1,0)$, $(1,1)$, and $(1,-1)$. Here, two different level curves intersect, showing us that there are directions where the value is increasing and directions where the value is decreasing.
$f_{x}=3-3x^2$ and $f_y=-4y+4y^3$.
When $x=-1$ and $x=1$, $f_{x}=3-3x^2=0$.
When $y=-1$, $y=0$, and $y=1$, $f_{y}=-4y+4y^3=0$.
Therefore we have the following critical points, $(-1, -1)$, $(-1, 0)$, $(-1, 1)$, $(1, -1)$, $(1, 0)$, and $(1, 1)$.
Let us now calculate the second partial derivatives.
$f_{xx}=-6x$, $f_{yy}=-4+12y^2$, and $f_{xy}=0$.
Let us now calculate the D values.
$D(-1,-1)=(6)(8)>0$ and $f_{xx}=6>0$, so $(-1,-1)$ is a local minimum.
$D(-1,0)=(6)(-4)<0$, so $(-1,0)$ is a saddle point.
$D(-1,1)=(6)(8)>0$ and $f_{xx}=6>0$, so $(-1,1)$ is a local minimum.
$D(1,-1)=(-6)(8)<0$, so $(1,-1)$ is a saddle point.
$D(1,0)=(-6)(-4)>0$ and $f_{xx}=-6<0$, so $(1,0)$ is a local maximum.
$D(1,1)=(-6)(8)<0$, so $(1,1)$ is a saddle point.
