ODE $y(1+2xy)dx+x(1-xy)dy=0$ 
$$y(1+2xy)dx+x(1-xy)dy=0$$

I have tried to isolate $\frac{dy}{dx}$ and got the following:
$$\frac{dy}{dx}=-\frac{y(1+2xy)}{x(1-xy)}$$
but I understand that the terms have to be in the same order and that is not the case. 
What should I do?

$$\frac{dy}{dx}=-\frac{y(1+2xy)}{x(1-xy)}$$
Let $v=yx$ so $\frac{dy}{dx}=x^{-1}\frac{dv}{dx}-x^{-2}v$
$$x^{-1}\frac{dv}{dx}-x^{-2}v=-\frac{v(1+2v)}{x^2(1-v)}$$
$$\frac{dv}{dx}=\frac{3v^2}{(v-1)x}$$
$$\frac{v-1}{3v^2}\frac{dv}{dx}=x^{-1}$$
You can then integrate both sides.
$$\frac13\left(v^{-1}+\log|v|\right)=\log|x|+\frac{c}{3}$$
$$\frac{1}{xy}+\log|y|+\log|x|=3\log|x|+c$$
$$\frac{1}{xy}+\log|y|-2\log|x|=c$$
To write this as a fuction of $y$ requires the Lambert W function.
