The correct option is C (x2+y2)2xy is a homogenous function of degree zero.
Remember always that the existence of a homogeneous function is not enough for the differential equation to be homogeneous. That homogeneous function should have degree 0 as well.
Now x2+y22xy. To check its degree replace x by kx and y by ky.
So it becomes k2x2+k2y22k2xy
=k22k2(x2+y2)xy=k02(x2+y2)xy
Since k0 is the common factor. So its degree is zero and this is the reason that given equation is a homogeneous differential equation.