Show that (x^2+3xy+y^2)dx-x^2*dy=0 is homogeneous.

Looking at (x^{2 }+ 3xy + y^{2})dx - x^{2}dy = 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**) = x^{2} + 3xy + y^{2} and N(**v**) = -x^{2}, we can establish homogeneity as follows.

M(a**v**) = (ax)^{2} + 3(ax)(ay) + (ay)^{2} = a^{2}x^{2} + 3a^{2}xy + a^{2}y^{2} = a^{2}(x^{2} + 3xy + y^{2}) = a^{2}M(**v**).

N(a**v**) = -(ax)^{2} = -a^{2}x^{2} = a^{2}N(**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.