Rational ODE $y'=\frac{3x^2-2xy+2}{6y^2-x^2+3}$ I really don't know how to start to solve this ODE:
$$y'=\frac{3x^2-2xy+2}{6y^2-x^2+3}$$
I know that somehow I have to isolate $y$ but how?
 A: When your ODE is such a mess, you can bet that it is exact or can be modified to be exact. 
In this case, you have
$$
3x^2-2xy+2+(6y^2-x^2+3)\frac{dy}{dx}=0,
$$
and 
$$
\frac{\partial}{\partial y}(3x^2-2xy+2)=-2x,\ \ \ \ \frac{\partial}{\partial x}(6y^2-x^2+3)=-2x.
$$
A: Hint:
Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=164:
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{3x^2-2x^2u+2}{6x^2u^2-x^2+3}$
$x\dfrac{du}{dx}=\dfrac{(3-2u)x^2+2}{(6u^2-1)x^2+3}-u$
$x\dfrac{du}{dx}=\dfrac{(3-u-6u^3)x^2-3u+2}{(6u^2-1)x^2+3}$
Let $v=x^2$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=2x\dfrac{du}{dv}$
$\therefore2x^2\dfrac{du}{dv}=\dfrac{(3-u-6u^3)x^2-3u+2}{(6u^2-1)x^2+3}$
$2v\dfrac{du}{dv}=\dfrac{(3-u-6u^3)v-3u+2}{(6u^2-1)v+3}$
$((3-u-6u^3)v-3u+2)\dfrac{dv}{du}=2(6u^2-1)v^2+6v$
This belongs to an Abel equation of the second kind.
