Answer
Local maximum: $f(0,0)=2$, Local minimum $f(0,4)=-30$, Saddle points $(2,2),(-2,2)$
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 $(x,y)=(2,2)$ and $(x,y)=(-2,2)$
$D(2,2)=-144 \lt 0$
$D(0,0)=144 \gt 0$ and $f_{xx}(0,0)\lt 0$ For $(x,y)=(0,4)$ $D(0,4)=144 \gt 0$ and $f_{xx}(0,0)\gt 0$
Hence, Local maximum: $f(0,0)=2$ Local minimum $f(0,4)=-30$ Saddle points $(2,2),(-2,2)$
