Official solution
\[{{xdy} \over {dx}} = y\left( {\log y - \log x + 1} \right)\]
\[{{dy} \over {dx}} = {y \over x}\left( {\log \left( {{y \over x}} \right) + 1} \right)\]
Put \[\,\,\,\,y = vx\]
\[{{dy} \over {dx}} = v + {{xdv} \over {dx}}\]
\[\,\,\,\,\,\,\,\,\,\,\, \Rightarrow v + {{xdv} \over {dx}} = v\left( {\log v + 1} \right)\]
\[{{xdv} \over {dx}} = v\,\log \,v\]
\[\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{dv} \over {v\,\log \,v}} = {{dx} \over x}\]
Put \[\,\,\,\,\log \,v = z\]
\[ \Rightarrow {1 \over v}dv = dz\]
\[ \Rightarrow {{dz} \over x} = {{dx} \over x}\]
\[ \Rightarrow \ln \,z = \ln x + \ln \,c\]
\[x = cx\,\,\,\,\] or \[\,\,\,\,\,\log v = cx\,\,\,\]
or \[\,\,\,\,\] \[\log \left( {{y \over x}} \right) = cx.\]