If $y={\tan }^{-1}\left(\sec x^3-\tan x^3\right),\frac{\pi }{2}<x^3<\frac{3\pi }{2}$, then

y $$\begin{gathered} ={\tan }^{-1}\left(\frac{1-\sin x^3}{\cos x^3}\right) \\ ={\tan }^{-1}\left(\frac{1-\cos \left(\frac{\pi }{2}-x^3\right)}{\sin \left(\frac{\pi }{2}-x^3\right)}\right) \\ ={\tan }^{-1}\left[\frac{2{\sin }^2\left(\frac{\pi }{4}-\frac{x^3}{2}\right)}{2\sin \left(\frac{\pi }{4}-\frac{x^3}{2}\right)\cos \left(\frac{\pi }{4}-\frac{x^3}{2}\right)}\right]={\tan }^{-1}\left(\tan \left(\frac{\pi }{4}-\frac{x^3}{2}\right)\right) \end{gathered}$$

$$\begin{gathered} y=\frac{\pi }{4}-\frac{x^3}{2} \left(\because -\frac{\pi }{2}<\frac{\pi }{4}-\frac{x^3}{2}<0\right) \\ =y^{\text{'}}=-\frac{3}{2}{x}^2 \\ =y^{\text{'}\text{'}}=-3x \end{gathered}$$

option (1) $x{y}^{\text{'}\text{'}}+2y^{\text{'}}=-3x^2+2\left(\frac{-3}{2}{x}^2\right)=-6x^2$ hence 1 is incorrect option $\left(2\right) x^2{y}^{\text{'}\text{'}}-6y+\frac{3\pi }{2}=x^2(-3x)-6\left(\frac{\pi }{4}-\frac{x^3}{2}\right)+\frac{3\pi }{2}=0$

 option 2 is correct 

similarly check other option