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

We are given $$y = \tan^{-1}(\sec x^3 - \tan x^3)$$ where $$\frac{\pi}{2} < x^3 < \frac{3\pi}{2}$$.

We use the identity: $$\sec\theta - \tan\theta = \frac{1 - \sin\theta}{\cos\theta}$$.

Recall that $$\frac{1 - \sin\theta}{\cos\theta} = \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right)$$.

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

Since $$\frac{\pi}{2} < x^3 < \frac{3\pi}{2}$$, we have $$\frac{\pi}{4} < \frac{x^3}{2} < \frac{3\pi}{4}$$, so $$\frac{\pi}{4} - \frac{x^3}{2} \in \left(-\frac{\pi}{2}, 0\right)$$, which lies within the principal range of $$\tan^{-1}$$.

Thus: $$y = \frac{\pi}{4} - \frac{x^3}{2}$$

Differentiating with respect to $$x$$:

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

Differentiating again:

$$y'' = -3x$$

Now we check Option B: $$x^2 y'' - 6y + \frac{3\pi}{2} = 0$$

Substituting:

$$x^2(-3x) - 6\left(\frac{\pi}{4} - \frac{x^3}{2}\right) + \frac{3\pi}{2}$$

$$= -3x^3 - \frac{6\pi}{4} + \frac{6x^3}{2} + \frac{3\pi}{2}$$

$$= -3x^3 - \frac{3\pi}{2} + 3x^3 + \frac{3\pi}{2}$$

$$= 0$$

The differential equation is satisfied.

The correct answer is Option B.