Find the general solution of the differential equation $\displaystyle 3y+x^3y^4+3xy'=0$.

$\displaystyle y'+\frac{y}{x}=\frac{-x^2y^4}{3}$

$\displaystyle xy'+y=\frac{-x^3y^4}{3}$

$\displaystyle \frac{d}{dx}(xy)=\frac{-x^3y^4}{3}$

$\displaystyle xy=\frac{-x^4y^4}{12}+C$

But the answer given in the book is $\displaystyle y(x)=x^{-1}(C+\ln x)^{-1/3}$. Where have I gone wrong?