Answer
Minima $f(x,y)=1$ at all the points (x,y) along $x-,y-$ axes.
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.
If we put either $x$ or $y$ equals $0$, we will get no specific or valid points.
For $(x,y)=( x,y)$
$D(x,y)\gt 0$ ; and $f_{xx}(x,y) \gt 0$; as per derivative test , that is, 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.
Also, $f(x,y)=xy+e^{-xy}=1$ at points $(0,0)$
Hence,
Minima $f(x,y)=1$ at all the points (x,y) along $x-,y-$ axes.
