Answer
$y = \tan x{\text{ is a solution}}$
Work Step by Step
$$\eqalign{
& y = \tan x,{\text{ }}y' - {y^2} = 1 \cr
& {\text{Differentiate the given function}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\tan x} \right] \cr
& \frac{{dy}}{{dx}} = {\sec ^2}x \cr
& or \cr
& y' = {\sec ^2}x \cr
& \cr
& {\text{Substitute }}y{\text{ and }}y'{\text{ into the given differential equation}} \cr
& \underbrace {y' - {y^2} = 1}_ \downarrow \cr
& {\sec ^2}x - {\left( {\tan x} \right)^2} = 1 \cr
& {\sec ^2}x - {\tan ^2}x = 1 \cr
& {\text{Use the identity ta}}{{\text{n}}^2}x = {\sec ^2}x - 1 \cr
& {\sec ^2}x - \left( {{{\sec }^2}x - 1} \right) = 1 \cr
& {\sec ^2}x - {\sec ^2}x + 1 = 1 \cr
& 1 = 1 \cr
& {\text{The statement is true}}{\text{, then }}y = \tan x{\text{ is a solution}} \cr
& {\text{of the given differential equation}}{\text{.}} \cr} $$
