Answer
There is a relative minimum at $(-2,-2)$
Work Step by Step
$f_{x}(x,y)=2x - y + 2 = 0$
$f_{y}(x,y)=-x + 2y+2= 0 \rightarrow x=2y+2$
Use the substitution method to solve the system of equations
$2(2y+2) - y + 2 =0 \rightarrow y=-2$
$x=-2$
The critical point is $(-2,-2)$
$f_{xx}(x,y)=2$
$f_{yy}(x,y)=2$
$f_{xy}(x,y)=-1$
$f_{xx}(-2,-2)=2$
$f_{yy}(-2,-2)=2$
$f_{xy}(-2,-2)=-1$
$D=2*2-(-1)^{2}=3$
Since there is D>0, and $f_{xx}(x,y)=2 \gt 0$, there is a relative minimum at $(-2,-2)$
