$\begin{aligned} & y=\tan ^{-1}\left(\sec x^3-\tan x^3\right) \\ & =\tan ^{-1}\left(\frac{1-\sin x^3}{\cos x^3}\right) \\ & =\tan ^{-1}\left(\frac{1-\cos \left(\frac{\pi}{2}-x^3\right)}{\sin \left(\frac{\pi}{2}-x^3\right)}\right) \\ & =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x^3}{2}\right)\right) \\ & \text { Since } \frac{\pi}{4}-\frac{x^3}{2} \in\left(-\frac{\pi}{2}, 0\right) \\ & y=\left(\frac{\pi}{4}-\frac{x^3}{2}\right) \\ & y^{\prime}=\frac{-3 x^2}{2}, y^{\prime \prime}=-3 x \\ & 4 y=\pi-2 x^3 \\ & 4 y=\pi-2 x^2\left(\frac{-y}{3}\right) \\ & 12 y=3 \pi+2 x^2 y^{\prime} \\ & 12 y=3 \pi+2 x^2 y^{\prime} \\ & x^2 y^{\prime \prime}-6 y+\frac{3 \pi}{2}=0\end{aligned}$