I want to see if I have the correct answer here.
I started by taking the gradient.
$$
\nabla f= \left \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right \rangle=\left \langle y-2,2y+x \right \rangle
$$
$y-2=0\Rightarrow y=2;2y+x=0\rightarrow 2(2)+x=0\Rightarrow x=-4$
This is the graph of the set D.
$L_1:y=0;f(x,0)=-2x, 0\leq x\leq 1$ $f_{min}=f(0,0)=0;f_{max}=f(1,0)=-2$
$L_2:x=1;f(1,y)=y^2+y-2,\sqrt{x}\leq y\leq 1\rightarrow 0\leq y\leq 1\Rightarrow f_{min}=f(1,0)=-2;f_{max}=f(1,1)=0$
$L_3:f(x,\sqrt{x})=(\sqrt{x})^2+x(\sqrt{x})-2x=x+x\sqrt{x}-2x;0\leq x\leq 1\Rightarrow f_{min}=f(0,0)=0;f_{max}=f(1,1)=0$
$f(2,-4)=4$ which is not on D
$f(1,0)=-2$
$f(0,0)=0$
$f(1,1)=0$
Which would imply local maximums at f(0,0) and f(1,1) and a local minimum at f(1,0) on D. Is this correct?
EDIT: Realized I'm considering the wrong region. Now I have the points
$f(2,-4)=4$ which is not on D
$f(1,1)=0$
$f(0,0)=0$
$f(0,1)=1$