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