How to solve this first-order ODE $\frac{dy}{dx}=\frac{y^6-2x^2}{2x^2y+2y^3-y}$? solve this ODE equation

$$\dfrac{dy}{dx}=\dfrac{y^6-2x^2}{2x^2y+2y^3-y}$$

My try:

$$\dfrac{y~dy}{dx}=\dfrac{y^6-2x^2}{2x^2+2y^2-1}$$
  let
  $$u=y^2$$
  then
  $$\dfrac{du}{dx}=\dfrac{2u^3-4x^2}{2x^2+2u-1}$$

then I can't work. Thank you, this problem is from ODE equation excise book,

and this book only take this answer:
$$(y^3-3x)^7(y^3+2x)^3=cx^{15}$$
 A: You did it correctly.
Now for $\dfrac{du}{dx}=\dfrac{2u^3-4x^2}{2x^2+2u-1}$ ,
$(2u+2x^2-1)\dfrac{du}{dx}=2u^3-4x^2$
Let $v=u+x^2-\dfrac{1}{2}$ ,
Then $u=v-x^2+\dfrac{1}{2}$
$\dfrac{du}{dx}=\dfrac{dv}{dx}-2x$
$\therefore2v\left(\dfrac{dv}{dx}-2x\right)=2\left(v-x^2+\dfrac{1}{2}\right)^3-4x^2$
$2v\dfrac{dv}{dx}-4xv=2v^3-6\left(x^2-\dfrac{1}{2}\right)v^2+6\left(x^2-\dfrac{1}{2}\right)^2v-2\left(x^2-\dfrac{1}{2}\right)^3-4x^2$
$2v\dfrac{dv}{dx}=2v^3-6\left(x^2-\dfrac{1}{2}\right)v^2+\left(6\left(x^2-\dfrac{1}{2}\right)^2+4x\right)v-2\left(x^2-\dfrac{1}{2}\right)^3-4x^2$
$v\dfrac{dv}{dx}=v^3-3\left(x^2-\dfrac{1}{2}\right)v^2+\left(3\left(x^2-\dfrac{1}{2}\right)^2+2x\right)v-\left(x^2-\dfrac{1}{2}\right)^3-2x^2$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $v=\dfrac{1}{w}$ ,
Then $\dfrac{dv}{dx}=-\dfrac{1}{w^2}\dfrac{dw}{dx}$
$\therefore-\dfrac{1}{w^3}\dfrac{dw}{dx}=\dfrac{1}{w^3}-3\left(x^2-\dfrac{1}{2}\right)\dfrac{1}{w^2}+\left(3\left(x^2-\dfrac{1}{2}\right)^2+2x\right)\dfrac{1}{w}-\left(x^2-\dfrac{1}{2}\right)^3-2x^2$
$\dfrac{dw}{dx}=\left(\left(x^2-\dfrac{1}{2}\right)^3+2x^2\right)w^3-\left(3\left(x^2-\dfrac{1}{2}\right)^2+2x\right)w^2+3\left(x^2-\dfrac{1}{2}\right)w-1$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
