Answer
Maximum value: $f(-1,0)=2$
Minimum value: $f(1,\pm 1)=-3$
Saddle point $(-1,\pm 1), (1,0)$
Work Step by Step
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point.
Critical point are: $(1,-1),(1,0),(1,1),(-1,-1),(-1,0),(-1,1)$
For $(x,y)=(- 1,0)$
$D=72xy^2-24x \gt 0$ ; and $f_{xx} =6x=-6\lt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
For $(x,y)=(1,1)$
$D=72xy^2-24x \gt 0$ ; and $f_{xx} =6x=6\gt 0$
For $(x,y)=(1,-1)$
$D=72xy^2-24x \gt 0$ ; and $f_{xx} =6x\gt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local minimum.
For $(x,y)=(-1,\pm 1)$ and $(1,0)$
$D=72xy^2-24x \lt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ is a saddle point.
Now, $f(-1,0)=x^3-3x+y^4-2y^2=2$
and
$f(1,\pm 1)=x^3-3x+y^4-2y^2=-3$
Therefore, we have Maximum value: $f(-1,0)=2$
Minimum value: $f(1,\pm 1)=-3$
Saddle point $(-1,\pm 1), (1,0)$
