Let y=vx, then y'=v+xv'.
(3y3-5yx2)dx+(xy2+x3)dy=0,
(3y3-5yx2)+(xy2+x3)y'=0,
(3v3x3-5vx3)+(v2x3+x3)(v+xv')=0,
3v3-5v+(v2+1)(v+xv')=0,
3v3-5v+v3+xv2v'+v+xv'=0,
4v3-4v+xv'(v2+1)=0,
x(dv/dx)(v2+1)=4v-4v3,
(v2+1)dv/(v-v3)=4dx/x.
Partial fractions:
(v2+1)/(v(1-v2))=A/v+B/(1-v)+C/(1+v),
v2+1≡A-Av2+Bv+Bv2+Cv-Cv2;
Equating coefficients of like terms:
A=1; -A+B-C=1, B-C=2; B+C=0, C=-B⇒2B=2, B=1, C=-1.
∫(v2+1)dv/(v-v3)=4∫dx/x becomes:
∫dv/v+∫dv/(1-v)-∫dv/(1+v)=4ln|ax| where a is a constant,
ln|v|-ln|1-v|-ln|1+v|=4ln|ax|=ln(a4x4)=ln(cx4) where c=a4 a positive constant.
v/(1-v2)=cx4, (y/x)/(1-y2/x2)=cx4, equating log arguments,
xy/(x2-y2)=cx4, y/(x2-y2)=cx3.
This can be written various ways: y=cx3(x2-y2)=cx5-cx3y2, cx3y2+y-cx5=0, ...
x=1, y=2 satisfies the equation:
2=-3c, c=-⅔, 3y+2x3(x2-y2)=0 or 2x5-2x3y2+3y=0.