I wonder if the ODE is supposed to be
[tex]\left(\ln x+\dfrac yx\right)\,\mathrm dx=-\ln x\,\mathrm dy[/tex]
(since having [tex]yx[/tex] makes this look increasingly more difficult...)
Rewrite the ODE as
[tex]\ln x\dfrac{\mathrm dy}{\mathrm dx}+\dfrac yx=-\ln x[/tex]
Notice that the left side is the derivative of a product:
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[y\ln x\right]=-\ln x[/tex]
Integrate both sides to get
[tex]y\ln x=x\ln x-x+C[/tex]
[tex]\implies\boxed{y(x)=x-\dfrac x{\ln x}+\dfrac C{\ln x}}[/tex]
