If $y={\tan }^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3/2}}\right)$  then, y''(1) is equal to

$y={\tan }^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3/2}}\right)={\tan }^{-1}\left(\frac{\sqrt{x}-x}{1+\sqrt{x}\cdot x}\right)$
$={\tan }^{-1}\left(\sqrt{x}\right)-{\tan }^{-1}\left(x\right)$
On differentiating w.r.t. X, we get
$\begin{array}{r}{y}^{\mathrm{\prime }}=\frac{1}{1+x}\cdot \frac{1}{2\sqrt{x}}-\frac{1}{1+x^2}\\ \Rightarrow y^{\mathrm{\prime }}\left(1\right)=\frac{1}{2}\cdot \frac{1}{2}-\frac{1}{2}=-\frac{1}{4}\end{array}$