Answer
Maximum value: $f(-1,0) =2$, Minimum value: $f(1,0) =-2$
Saddle point: $(0, \pm 1)$
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 not a local minimum and local maximum or, a saddle point.
For $(x,y)=( -1,0)$
$D(-1,0)=36 \gt 0$ ; and $f_{xx}(-1,0)=6 \gt 0$
For $(x,y)=(1,0)$
$D(1,0)=36 \gt 0$ ; and $f_{xx}(-1,0)=-6 \lt 0$
$D(0, \pm 1)=-36 \lt 0$; saddle points
Hence,
Maximum value: $f(-1,0) =2$, Minimum value: $f(1,0) =-2$
Saddle point:$(0, \pm 1)$
