Answer
$$y' =\frac{1}{2\sqrt{x}\left((\sin^{-1}\sqrt{x})^2+1\right)\sqrt{1-x}}$$
Work Step by Step
Given $$y=\tan^{-1}\left(\sin^{-1}\sqrt{x}\right)$$
Then
\begin{align*}
y'&=\frac{1}{1+(\sin^{-1}\sqrt{x})^2}\frac{1}{\sqrt{1-x}}\frac{1}{2\sqrt{x}}\\
&=\frac{1}{2\sqrt{x}\left((\sin^{-1}\sqrt{x})^2+1\right)\sqrt{1-x}}
\end{align*}