Answer
$y = \ln x{\text{ is not a solution}}$
Work Step by Step
$$\eqalign{
& y = \ln x,{\text{ }}xy'' - y' = 0 \cr
& {\text{Differentiate the given function}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln x} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{x} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {\frac{1}{x}} \right] = - \frac{1}{{{x^2}}} \cr
& or \cr
& y' = \frac{1}{x}{\text{ and }}y'' = - \frac{1}{{{x^2}}} \cr
& \cr
& {\text{Substitute }}y''{\text{ and }}y'{\text{ into the given differential equation}} \cr
& \underbrace {xy'' - y' = 0}_ \downarrow \cr
& x\left( { - \frac{1}{{{x^2}}}} \right) - \frac{1}{x} = 0 \cr
& - \frac{1}{x} - \frac{1}{x} = 0 \cr
& - \frac{2}{x} \ne 0 \cr
& {\text{The statement is false}}{\text{, then }}y = \ln x{\text{ is not a solution}} \cr
& {\text{of the given differential equation}}{\text{.}} \cr} $$
