The function is continuous and differentiable, so its maximum value over a region will be critical points or at the boundary.
To find critical points, we find $f_x$ and $f_y$ and set them both equal to $0$: $f_x = 2x+y = 0$ $f_y = 2y+x = 0$ $2(-2x)+x = 0$ $x = 0, y = 0$
So the origin is a critical point. Using the second derivative test in two dimensions: $f_{xx} = 2$ $f_{xy} = f_{yx} = 1$ $f_{yy} = 2$ Because $f_{xx}*f_{yy} - f_{xy}^2 = 3 > 0$ and $f_{xx}*f_{yy} > 0$ (so its a min, not a max), $(0,0)$ is a minimum, so it cannot possible be the greatest value.
Now we have to test the boundary, as the greatest value has to be there. Our new constraint is $g(x,y) = x^2+y^2 = 1$, so $\nabla g = <2x, 2y>$
Using Lagrange multipliers, we know $\nabla f = \lambda \nabla g$: $2x+y = \lambda 2x$ $2y +x = \lambda 2y$
${2x+y \over 2x} = {2y+x \over 2y}$ $4xy+2y^2 = 4xy + 2x^2$ $x^2 = y^2$ $x = \pm y$ Using our constraint, we have $x^2 +x^2 = 1$, so $x = \pm 1/ \sqrt{2}$. So, our possible maximum values are $(1/ \sqrt{2}, 1/ \sqrt{2}), (-1/ \sqrt{2}, 1/ \sqrt{2}), (1/ \sqrt{2}, -1/ \sqrt{2}), (-1/ \sqrt{2}, -1/ \sqrt{2})$. After checking values of $f$ at each of these points, we can conclude that the greatest $f$ value, $3/2$, occurs at $(1/ \sqrt{2}, 1/ \sqrt{2})$ and $(-1/ \sqrt{2}, -1/ \sqrt{2})$.