Let \(v(x)=xy(x)\), so that \(\dfrac yx=\dfrac v{x^2}\), and \(\dfrac{\mathrm dv}{\mathrm dx}=x\dfrac{\mathrm dy}{\mathrm dx}+y\), or \(\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dv}{\mathrm dx}-\dfrac v{x^2}\).
So the ODE is
\((1+yx)x\,\mathrm dy+(1-yx)y\,\mathrm dx=0\iff\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{(1-yx)y}{(1+yx)x}\)
\(\implies \dfrac1x\dfrac{\mathrm dv}{\mathrm dx}-\dfrac v{x^2}=-\dfrac{1-v}{1+v}\dfrac v{x^2}\)
\(\implies \dfrac{\mathrm dv}{\mathrm dx}=\dfrac vx\left(1-\dfrac{1-v}{1+v}\right)\)
\(\implies\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v}{x(1+v)}\)
This ODE is separable, so you can write
\(\dfrac{1+v}{2v}\,\mathrm dv=\dfrac{\mathrm dx}x\)
and integrating both sides yields
\(\displaystyle\int\frac{1+v}{2v}\,\mathrm dv=\int \dfrac{\mathrm dx}x\)
\(\dfrac12\left(v+\ln|v|\right)=\ln|x|+C\)
\(v+\ln|v|=2\ln|x|+C\)
Replace \(v=xy\) and you end up with
\(xy-\ln|xy|=2\ln|x|+C\)
