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