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