#### Answer

$$y=\sin^{-1}x$$

#### Work Step by Step

Given $$y'= \sec y$$
Then
\begin{align*}
\frac{dy}{dx}&= \sec y\\
\frac{1}{\sec y} dy&=dx\\
\int \cos y dy&=\int dx\\
\sin y&=x+C
\end{align*}
Since $y(0)=0$, then $C=0$, and hence
$$ \sin y=x\ \ \Rightarrow \ \ y=\sin^{-1}x$$