$y = Ax:xy' = y(x \ne 0)$

.: We have, $y = Ax$

…(1)
Differentiating (1) w.r.t. $x$,

we get $y' = A$

…(2)
Dividing (2) by (1),

we get $\cfrac{{y'}}{y} = \cfrac{1}{x} \Rightarrow xy' = y$

Hence, $y = Ax$ is a solution of the given differential equation.