Completing the square in the expression for f(x,y) gives . From this (or by taking partial derivatives), you see that the only critical point of f(x,y) is a global minimum value of 0, at the origin. The origin is inside D, so the minimum value of f(x,y) in D is 0.
To find the maximum value of f in D, you need to check what happens on the boundary of D, in other words when f satisfies the constraint . You do this by the Lagrange multiplier method, and apart from the error of having a 2 that should be 4 you seem to have done this correctly. So the maximum value of f in D is 2, at the points .