TITLE: Solving $(y+x^4y^2)dx+xdy=0$
QUESTION [6 upvotes]: How to solve (in terms of $y$) $(y+x^4y^2)dx+xdy=0$.
I know I'm supposed to multiply by an integrating factor to turn this equation into an exact equation.
In the previous exercise I proved that in $Mdx+Ndy$ the functions:

$\frac{1}{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)$
$\frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)$
If $M=yf(xy)$ and $N=xg(xy)$ then $\frac{1}{xM-yN}$ is an integrating factor

I've tried using 1. and 2., but the calculations are horrible and do not seem to work. I don't think that's the way to go.
About 3. I don't know what $f$ and $g$ I should choose.
how can I go about solving this differential equation?
this isn't homework.

REPLY [3 votes]: An integrating factor, as you can check (!), is
$$f(x,y) = \frac1{xy(x^4y-3)}\,.$$
Where does this come from, you ask? By noting that the $1$-parameter group on $\mathbb R^2-\{0\}$
$$\{(x,y) \rightsquigarrow (e^t x, e^{-4t}y)\}$$
leaves the differential equation invariant. I was told years ago that it was exactly for the purposes of finding such obscure integrating factors that Sophus Lie "invented" Lie groups.
One then integrates and gets 
$$\frac{3-x^4y}{xy} = C,$$
which is, indeed, $y=\dfrac3{x^4+Cx}$, as @Amzoti obtained.