Answer
Local maximum:$(1,1),(-1,-1)$;
Local minimum: $(-1,1),(1,-1)$;
Saddle point: $(0,0)$
Work Step by Step
Given: $f(x,y)=xye^{-(x^2+y^2)/2}$
To solve this problem we will take the help of Second derivative test that suggests the following conditions to determine the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$.
i) 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.
ii) 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.
iii) 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.
For $(x,y)=(1,1)$, we have $D(1,1) \gt 0$ and $f_{xx}(0,0) \lt 0$; For $(x,y)=(-1,1)$, we have $D(-1,1) \gt 0$ and $f_{xx}(0,0) \gt 0$
For $(x,y)=(0,0)$: $D \lt 0$ $\implies$ saddle point.
This shows that
Local maximum:$(1,1),(-1,-1)$;
Local minimum: $(-1,1),(1,-1)$;
Saddle point: $(0,0)$
