Answer
Maximum value: $f(0,-1) =2$, Minimum value: $f(\pm 1,1) =-3$
Saddle point:$(0,1),(\pm 1,-1)$
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.
For $(x,y)=( 0,-1)$
$D(0,-1)=72 \gt 0$ ; and $f_{xx}(0,-1) \gt 0$
For $(x,y)=(\pm 1,1)$
$D(\pm 1,-1)=72 \gt 0$ ; and $f_{xx}(\pm 1,-1) \lt 0$
$D(0,1)=-72 \lt 0$; saddle points and $D(\pm 1,-1)=-72 \lt 0$; saddle points
Hence,
Maximum value: $f(0,-1) =2$, Minimum value: $f(\pm 1,1) =-3$
Saddle point:$(0,1),(\pm 1,-1)$
