Let #y=vx#, where v is a function of x, which means #dy/dx= v +x (dv)/dx#. The given DE , thus transforms to # v+x(dv)/dx =(v^3 +3v)/(1+3v^2)#

#x(dv)/dx = (v^3+3v)/(1+3v^2) -v =(2v-2v^3)/(1+3v^2)#

#(1+3v^2)/(v(1-v^2))dv= 2(1/x) dx#. Partial fractions on the left side would be

#(1/v +(4v)/(1-v^2))dv= 2 (1/x) dx#. Now integrate on both sides to get

#ln v -2ln(1-v^2) = 2 lnx +C#. On simplification it becomes,

#ln (v/(1-v^2)^2) = ln x^2 +C#

#v/(1-v^2)^2 = c_1x^2 #

#(yx^3)/(x^2-y^2)^2= c_1x^2 #

#(xy)/(x^2-y^2)^2 =c_1#