If $y = \tan^{-1}\left(\sec x^3 - \tan x^3\right), \frac{\pi}{2} < x^3 < \frac{3\pi}{2}$, then:
\( xy''+2y' = 0 \)
\( x^{2}y''-6y+\frac{3\pi}{2}=0 \)
\( x^{2}y''-6y+3\pi=0 \)
\( xy''-4y'=0 \)
