Answer
Minimum value: $f(1, \pm 1)=f(-1, \pm 1)=3$
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 $(x,y)=(1, \pm 1)$
$D=64-16=48\gt 0$ ; and $f_{xx} =8\gt 0$
For $(x,y)=(-1, \pm 1)$
$D=48\gt 0$ ; and $f_{xx}=8 \gt 0$
and $f_{xx}(x,y)f_{yy}(x,y)-[f_{xy}(x,y)]^2 = 48 \gt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
Therefore, we have Minimum value: $f(1, \pm 1)=f(-1, \pm 1)=3$
