Answer
Local maximum at $f(-2,-2)=4$
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, local maximum, or a saddle point.
From the given question, we have $f_x(x,y)=y-2-2x$ and $f_y(x,y)=x-2-2y$ gives $(x,y)=(-2,-2)$.
For $(x,y)=(-2,-2)$ $D(-2,-2)=3 \gt 0$ and $f_{xx}(-2,-2)=2 \lt 0$
This implies that there is a local maximum at $f(-2,-2)=4$
