Ordinary Differential Equation getting two different answers Solve $y(y^2-2x^2)dx+x(2y^2-x^2)dy=0$ and find a particular curve passing through $(1,2)$
My attempt:
1st Solution: Rewrite as
$y(y^2dx+x2ydy)-x(x^2dy+y2xdx)=0$
$\implies xy^2d(xy^2)-x^2yd(x^2y)=0$    (multiplying $xy$)
$\implies (xy^2)^2-(x^2y)^2 = c\,\,\,$   Or $x^2y^2(y^2-x^2) = c$
$\implies x^2y^2(y^2-x^2) = 12 $
2nd Solution: Rewrite as
$\frac{dy}{dx}=\frac{2\frac yx-(\frac yx)^3}{2(\frac yx)^2-1}$
Let $y=tx\implies \frac{dy}{dx}=t+x\frac{dt}{dx}=\frac{2t-t^3}{2t^2-1}$
$\implies x\frac{dt}{dx}=\frac{3(t-t^3)}{2t^2-1}$
$\implies \int\frac{2t^2-1}{t^3-t}dt+\int\frac 3x dx=0$
$\implies \int \left(\frac 2t+\frac{1}{(t-1)}+\frac{1}{(t+1)}\right)dt+\int\frac 6x dx = 0 $
$\implies \ln|t^2(t^2-1)x^6| = ln c $
$\implies |t^2(t^2-1)x^6| = c $
$\implies |x^2y^2(y^2-x^2)| = 12 $
$\implies x^2y^2(y^2-x^2) = \pm 12 $
Where I am doing wrong in second or in first solution!

In your second solution, you work out your constant too soon.  It is the case that if $|x| = c$, then you can conclude that $ x = c$ because the arbitrary constant can absorb the $\pm$.  In the first equation, $c$ nonzero.  In the second, it's allowed to be negative.  
So I think you should have written 
$$x^2y^2(y^2-x^2) = c$$ 
and THEN worked out what $c$ is.  
You'd have the same trouble if solved a DE and got $(y+x)^2 =c$ and then worked out that $c =4$ and then wrote $y+x = \pm 2$, when you should rather have done this:
$$(y+x)^2 = c$$
$$y+x = c$$
$$y+x = 2$$
