Answer
Minimum $f(0,-0.794) \approx -1.191,f(\pm 1.592,1.267) \approx -1.310$,
Saddle points at $(\pm 0.720,0.259)$
Lowest points: $(\pm 1.592,1.267, -1.310)$
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.
Critical points are: $(0,-0.794), (\pm 1.592,1.267) , (\pm 0.720,0.259),(\pm 1.592,1.267, -1.310)$
For $(x,y)=f(0,-0.794)$
$D(0,-0.794)=4 \gt 0$ and $f_{xx}=2 \gt 0$
For $(x,y)=f(\pm 1.592,1.267) $
$D(\pm 1.592,1.267) =-2 \gt 0$ and $f_{xx}=2 \gt 0$
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.
For $(x,y)=(\pm 0.720,0.259)$
$D(\pm 0.720,0.259) =-8 \lt 0$ ; saddle points.
Hence,
Minimum $f(0,-0.794) \approx -1.191,f(\pm 1.592,1.267) \approx -1.310$,
Saddle points at $(\pm 0.720,0.259)$
Lowest points: $(\pm 1.592,1.267, -1.310)$
