Answer
Maximum value: $f(1,1)=1$
Saddle point $(0,0),(0,3),(3,0)$
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 a not a local minimum and local maximum or, a saddle point.
Critical point are: $(1,1),(0,0),(0,3),(3,0)$
For $(x,y)=(1,1)$
$D=3 \gt 0$ ; and $f_{xx} =-2\gt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local maximum.
For $(x,y)=(0,0)$
$D=-9 \lt 0$
For $(x,y)=(0,3)$
$D=-9 \lt 0$
For $(x,y)=(3,0)$
$D=-9 \lt 0$
When$D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point.
Now, $f(1,1)=3xy-x^2y-xy^2=3(1)(1)-(1)^2(1)-(1)(1)^2=1$
Therefore, we have Maximum value: $f(1,1)=1$
Saddle point $(0,0),(0,3),(3,0)$
