Answer
Maximum value: $f(-1,0) =5$, Minimum value: $f(3,2) =-31$
Saddle point:$f(3,0)=27,f(-1,2)=7$
Work Step by Step
Second derivative test: some noteworthy points to calculate the local minimum, local maximum and saddle points 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 not a local minimum and local maximum or, a saddle point.
For $(x,y)=( -1,0)$
$D(-1,0)=72 \gt 0$ ; and $f_{xx}(-1,0) \gt 0$
For $(x,y)=(3,2)$
$D(1,0)=72 \gt 0$ ; and $f_{xx}(-1,0) \lt 0$
$D(3,0)=-72 \lt 0$; saddle point and $D(-1,2)=-72 \lt 0$; saddle point
Hence,
Maximum value: $f(-1,0) =5$, Minimum value: $f(3,2) =-31$
Saddle point:$f(3,0)=27,f(-1,2)=7$
