Solve $xy\frac{dy}{dx} = \sqrt{x^2 - y^2 - x^2y^2 - 1}$ Can anyone please help me solving the below
$$
xy\frac{dy}{dx} = \sqrt{x^2 - y^2 - x^2y^2 - 1}
$$
If there would have been + instead of -, It would have been easier.
I tried 2 ways as mentioned below :
1. Putting $x^2 = u, y^2 = v$
2. Squaring both sides.Then, dividing both sides by $x^2y^2$
But didn't able to go much further.
Any hint would be appreciated. Thanks in advance.  

Hint:
Let $u=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=2x\dfrac{dy}{du}$
$\therefore2x^2y\dfrac{dy}{du}=\sqrt{u-y^2-uy^2-1}$
$2uy\dfrac{dy}{du}=\sqrt{u-y^2-uy^2-1}$
Let $v=y^2$ ,
Then $\dfrac{dv}{du}=2y\dfrac{dy}{du}$
$\therefore u\dfrac{dv}{du}=\sqrt{u-v-uv-1}$
Let $w=\sqrt{u-v-uv-1}$ ,
Then $w^2=u-1-(u+1)v$
$v=\dfrac{u-1}{u+1}-\dfrac{w^2}{u+1}$
$\dfrac{dv}{du}=\dfrac{2}{(u+1)^2}+\dfrac{w^2}{(u+1)^2}-\dfrac{2w}{u+1}\dfrac{dw}{du}$
$\therefore\dfrac{2u}{(u+1)^2}+\dfrac{uw^2}{(u+1)^2}-\dfrac{2uw}{u+1}\dfrac{dw}{du}=w$
$\dfrac{2uw}{u+1}\dfrac{dw}{du}=\dfrac{uw^2}{(u+1)^2}-w+\dfrac{2u}{(u+1)^2}$
$w\dfrac{dw}{du}=\dfrac{w^2}{2(u+1)}-\dfrac{(u+1)w}{2u}+\dfrac{1}{u+1}$
This belongs to an Abel equation of the second kind.
