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