(x²+2y²)+(4xy-y²)(dy/dx)=0.
Let y=vx then dy/dx=v+xdv/dx and:
(x²+2v²x²)+(4vx²-v²x²)(dy/dx)=0,
(1+2v²)+(4v-v²)(v+xdv/dx)=0 (dividing through by x²≠0),
xdv/dx=-(1+2v²)/(4v-v²)-v=(v³-6v²-1)/(4v-v²).
Separating variables:
(v(v-4)/(v³-6v²-1))dv=-dx/x.
Let p=v³-6v²-1, then dp=3v²-12v=[3v(v-4)]dv,
v(v-4)dv=dp/3.
So:
dp/(3p)=-dx/x and, integrating: ⅓ln(p)=ln(a/x) where a is constant of integration.
ln(p)=3ln(a/x)=ln(a/x)³.
Therefore, p=b/x³ where b=a³.
Back substitution:
v³-6v²-1=b/x³,
(y/x)³-6(y/x)²-1=b/x³,
y³-6xy²-x³=b.