Solve $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{(1+\log x+\log y)^2}$ Solve $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{(1+\log x+\log y)^2}$


I tried solving this differential equation, I got stuck after a few steps
I got 
$(a+\log(xy))^2d(xy) = xdx$
I don't know what to do after this
If you write $$\frac{dy}{dx}+\frac{y}{x} = \frac{1}{(1+\log(xy))^2}$$ and substitute $v=xy$, then wonderful things happen.  Then $$\frac{dv}{dx} = x\frac{dy}{dx} +y$$ Plug stuff in and the equation becomes $$\frac{1}{x}\frac{dv}{dx}=\frac{1}{(1+\log v)^2},$$ separable.
With $u = xy$, the differential equation becomes
separable.