Let y=vx, then dy/dx=v+xdv/dx.
The DE becomes:
vx(x2+v2x2)+x(3x2-5v2x2)(v+xdv/dx)=0,
v(1+v2)+(3-5v2)(v+xdv/dx)=0,
v+v3+3v+3xdv/dx-5v3-5v2xdv/dx=0,
4v-4v3=-x(3-5v2)dv/dx,
∫dx/x=¼∫[(5v2-3)/(v-v3)]dv.
The variables are now separated so integration should be easier.
The right hand integral can be split into partial fractions: -3/v+2v/(1-v2) which is easily integrated. So:
ln|x|=-3ln|v|-ln|1-v2|+A where A is a constant.
ln|x|=-ln|v3(1-v2)|+A. If A=ln(C) where C is a constant then:
ln|x|=ln(C/(v3-v4)) and x=C/(v3-v4).
But v=y/x, so x=C/((y/x)3-(y/x))=Cx3/(y3-x2y), Cx2=y3-x2y.
Initial conditions are (x,y)=(1,1) so C=0, y2=x2.
Now check out the original DE using dy/dx=x/y.
y2(x2+y2)+x2(3x2-5y2)=0, 2x4-2x4=0, which is true. So this confirms the solution y2=x2.