Answer
Saddle point at $(1,1),(-1,-1)$. No local minimum or maximum.
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, local maximum or a saddle point.
For the given question, we have $f_x(x,y)=1-2xy+y^2$ and $f_y(x,y)=-1-x^2+2xy$ gives $(x,y)=(1,1),(-1,-1)$.
For $(x,y)=(1,1)$
$D(1,1)=-4 \lt 0$ and $D(-1,-1)=-4 \lt 0$
This implies that there is no local minimum and no local maximum. There is a saddle point at $(1,1),(-1,-1)$.
