 Solution

For fixed boundary values, integration by parts gives $L_u-{\partial }_x{L}_{u_x}-{\partial }_y{L}_{u_y}=0$. For $L=u^2+u_x{u}_y$, this is $2u-2u_{xy}=0$, or $u_{xy}=u$, with the independently prescribed boundary data $u=xy$ on the circle.