Answer
Maximum value: $f(0,\pm 1)=\dfrac{2}{e}$ or, Maximum value: $f(0,\pm 1)=2e^{-1}$
Minimum value: $f(0,0)=0$
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.
Critical point are: $(0,\pm 1),(0,0)$
For $(x,y)=(0,\pm 1)$
$D=e^{-4}(4_y^2) \gt 0$ ; and $f_{xx} =-4\lt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local maximum.
For $(x,y)=(0,0)$
$D=0\gt 0$ ; and $f_{xx} \gt 0$
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.
Now, $f(0, 1)=e^{-x^2-y^2}(x^2+2y^2)=e^{-0-(1)^2}(0+2(1)^2)=2e^{-1}$
and
$f(0, -1)=e^{-x^2-y^2}(x^2+2y^2)=e^{-0-(-1)^2}(0+2(-1)^2)=2e^{-1}$
Therefore, we have Maximum value: $f(0,\pm 1)=\dfrac{2}{e}$ or, Maximum value: $f(0,\pm 1)=2e^{-1}$
Minimum value: $f(0,0)=0$
