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?

\begin{align*}y(1+2xy)\ dx+x(1−xy)\ dy=0&\implies (y\ dx+x\ dy)+2xy^2\ dx-x^2y\ dy=0\\&\implies d(xy)=-xy(2y\ dx-x\ dy)\\&\implies \dfrac{d(xy)}{xy}=-2y\ dx+x\ dy\\&\implies\dfrac{d(xy)}{x^2y^2}=-2\dfrac{dx}{x}+\dfrac{dy}{y}\end{align*}I think you can take it from here.
