Solution of the following differential equation . $2x^3ydy + (1-y^2)(x^2y^2 -1)dx =0$ 
The correct solution to the problem is :
$x^2y^2 = (Cx - 1)(1 - y^2)$ 
Where C is a constant .
 A: Let $$y(x)=\sqrt{v(x)}$$ then
$$v'(x)=\frac{v(x)^2}{x}+\frac{(-x^2-1)v(x)}{x^3}+\frac{1}{x^3}$$
Now let $$v(x)=u(x)$$ so
$$u'(x)=u(x)^2+\left(-\frac{1}{x}+\frac{-x^2-1}{x^3}\right)u(x)+\frac{1}{x^4}$$
with $$u(x)=-\frac{\frac{dw(x)}{dx}}{w(x)}$$ so
$$-w''(x)=-w'(x)\left(-\frac{1}{x}+\frac{-x^2-1}{x^3}\right)+\frac{w(x)}{x^4}$$
Let
$$w(x)=\frac{f(x)}{x}$$
so $$\frac{f''(x)}{x}+\frac {f'(x)}{x^4}=0$$
with $$f'(x)=w(x)$$
we get
$$w'(x)=\frac{w(x)}{x^3}$$
and
$$\int\frac{\frac{dw(x)}{dx}}{w(x)}dx=-\int \frac{1}{x^2}dx$$
A: This answer only shows that something is not working in the whole puzzle, i am not solving the given differential equation, but having the solution, we get an other differential equation.
The relation $x^2y^2 = (Cx - 1)(1 - y^2)$ may be rewritten as:
$$
\underbrace{
\frac 1x
\left(
\frac {x^2y^2}{1-y^2}+1
\right)}_{F=F(x,y)}
=C\ .
$$
Differentiating, $F'_x\; dx+F'_y\; dy=0$, does not deliver the given equation.
In fact, after multiplication of the partial $F'_x$, and $F'_y$ with the "common denominator" $G=x^2(1-y^2)^2$ we get the following values for $G\cdot F'_x$ and $G\cdot F'_y$:
sage: F = ( x^2*y^2 /(1-y^2) +1 ) / x
sage: G = x^2 * (y^2 - 1)^2
sage: ( G * diff(F, y) ).factor(), ( G * diff(F, x) ).factor()
(2*x^3*y, -(x^2*y^2 + y^2 - 1)*(y + 1)*(y - 1))

and the factor $(x^2y^2 + y^2 - 1)$ above in the $dx$-component is not matching the one in the given differential equation.
