Show that (x^2+3xy+y^2)dx-x^2*dy=0 is homogeneous.
Looking at (x2 + 3xy + y2)dx - x2dy = 0, note that is is already of the M(x,y)dx + N(x,y)dy.
Let v = <x,y>.
For the two functions M(v) = x2 + 3xy + y2 and N(v) = -x2, we can establish homogeneity as follows.
M(av) = (ax)2 + 3(ax)(ay) + (ay)2 = a2x2 + 3a2xy + a2y2 = a2(x2 + 3xy + y2) = a2M(v).
N(av) = -(ax)2 = -a2x2 = a2N(v).
Since M and N are the same degree, this same scalar multiplication property holds for the whole equation.
Thus M and N are homogeneous and so is your equation.