If for the differential equation $$y' = \frac{y}{x} + \phi \left( {\frac{x}{y}} \right),$$     the general solution is $$y = \frac{x}{{\log \left| {Cx} \right|}}$$   then $$\phi \left( {\frac{x}{y}} \right)$$  is given by :

Solution :
$$\eqalign{ & {\text{Putting }}v = \frac{y}{x}{\text{ so that }}x\frac{{dv}}{{dx}} + v = \frac{{dy}}{{dx}} \cr & {\text{We have }}x\frac{{dv}}{{dx}} + v = v + \phi \left( {\frac{1}{v}} \right) \cr & \Rightarrow \frac{{dv}}{{\phi \left( {\frac{1}{v}} \right)}} = \frac{{dx}}{x}\,; \cr & \Rightarrow \log \left| {Cx} \right| = \int {\frac{{dv}}{{\phi \left( {\frac{1}{v}} \right)}}} \cr & {\text{But }}y = \frac{x}{{\log \left| {Cx} \right|}}{\text{ is the general solution,}} \cr & {\text{So, }}\frac{x}{y} = \frac{1}{v} = \log \left| {Cx} \right| = \int {\frac{{dv}}{{\phi \left( {\frac{1}{v}} \right)}}} \cr & \Rightarrow \phi \left( {\frac{1}{v}} \right) = - \frac{1}{{{v^2}}}\,\,\,\,\left( {{\text{differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}v{\text{ both sides}}} \right) \cr & \Rightarrow \phi \left( {\frac{x}{y}} \right) = - \frac{{{y^2}}}{{{x^2}}} \cr} $$