Answer
Local minimum of $f(1,0)=0$
Work Step by Step
Since, we have $f_x(x,y)=2x-2y-2=0, f_y(x,y)=-2x+4y+2=0$
After solving the above two equations, we get
$x=1,y=0$
Thus the critical point is: $(1,0)$
As per second derivative test, we have
$D=f_{xx}f_{yy}-f^2_{xy}=4$ and $D=4 \gt 0$ and $f_{xx}=2 \gt 0$
Thus, local minimum of $f(1,0)=0$
