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

Given,

$y={\tan }^{-1}\left(\sec x^3-\tan x^3\right)$

$={\tan }^{-1}\left(\frac{1-\sin x^3}{\cos x^3}\right)$

$={\tan }^{-1}\left(\frac{1-\cos \left(\frac{\pi }{2}-x^3\right)}{\sin \left(\frac{\pi }{2}-x^3\right)}\right)$

$={\tan }^{-1}\left(\tan \left(\frac{\pi }{4}-\frac{x^3}{2}\right)\right)$

Since $\frac{\pi }{4}-\frac{x^3}{2}\in \left(-\frac{\pi }{2},0\right)$ as $\left(\frac{\pi }{2}<x^3<\frac{3\pi }{2}\right)$

So, $y=\left(\frac{\pi }{4}-\frac{x^3}{2}\right)$

Now differentiating we get,

$y^{\text{'}}=\frac{-3x^2}{2},y^{\text{'}\text{'}}=-3x$

Now putting the value of $x$ in term of $y^{\text{'}\text{'}}$ in $4y=\pi -2x^3$

We get,

$4y=\pi -2x^2\left(\frac{-y^{\text{'}\text{'}}}{3}\right)$

$12y=3\pi +2x^2{y}^{\text{'}\text{'}}$

$x^2{y}^{\text{'}\text{'}}-6y+\frac{3\pi }{2}=0$