I read this as:
dy/dx-(1/3x)y=y⁴ln(x), dy/dx=(1/3x)y+y⁴ln(x).
Let z=1/y³ (Bernoulli solution method),
dz/dx=-(3/y⁴)dy/dx=-(3/y⁴)((1/3x)y+y⁴ln(x)).
dz/dx=-1/(xy³)-3ln(x)=-z/x-3ln(x),
dz/dx+z/x=-3ln(x).
Multiply through by x:
xdz/dx+z=-3xln(x),
∫d(xz)=-3∫xln(x)dx+C.
xz=-3∫xln(x)dx+C.
Integrate by parts ∫xln(x)dx:
u=ln(x), dv=xdx,
du=dx/x, v=x²/2, so
∫xln(x)dx=x²ln(x)/2-∫(x²/2)(dx/x),
∫xln(x)dx=x²ln(x)/2-½∫xdx,
∫xln(x)dx=x²ln(x)/2-x²/4=x²ln(x²)/4-x²/4,
∫xln(x)dx=¼x²(ln(x²)-1).
xz=-¾x²(ln(x²)-1)+C
Substituting for z:
x/y³=-¾x²(ln(x²)-1)+C,
x=-¾x²y³(ln(x²)-1)+Cy³
Cy³-¾x²y³(ln(x²)-1)-x=0.
ay³-3x²y³(ln(x²)-1)-4x=0 or 3x²y³(ln(x²)-1)+4x-ay³=0 where a is a constant (a=4C).