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