1. Given function

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

Let $t=x^3$ Then $y=\tan^{-1}(\sec t-\tan t)$ with $\frac{\pi}{2}<t<\frac{3\pi}{2}$

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2. Simplify the expression inside

Use the identity: $\sec t-\tan t=\frac{1}{\sec t+\tan t}$ Also, $(\sec t-\tan t)(\sec t+\tan t)=\sec^2 t-\tan^2 t=1$

Now use the standard identity: $\sec t-\tan t=\tan\left(\frac{\pi}{4}-\frac{t}{2}\right)$

So, $y=\tan^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{t}{2}\right)\right)$

But we must choose the correct branch of $\tan^{-1}$, whose principal value lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

Now since $\frac{\pi}{2}<t<\frac{3\pi}{2}$ we get $\frac{\pi}{4}-\frac{t}{2}\in\left(-\frac{\pi}{2},0\right)$ which lies completely inside the principal branch.

Hence, $y=\frac{\pi}{4}-\frac{t}{2}=\frac{\pi}{4}-\frac{x^3}{2}$

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3. Differentiate

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

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

Second derivative: $y''=-3x$

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4. Check each option

Option A: $xy''+2y'=0$

Compute: $xy''+2y'=x(-3x)+2\left(-\frac{3x^2}{2}\right)=-3x^2-3x^2=-6x^2\neq 0$ So A is false.

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

Compute: $x^2y''=x^2(-3x)=-3x^3$

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

Therefore, $x^2y''-6y+\frac{3\pi}{2}=-3x^3+\left(-\frac{3\pi}{2}+3x^3\right)+\frac{3\pi}{2}=0$ So B is true.

Option C: $x^2y''-6y+3\pi=0$

From above, $x^2y''-6y+3\pi=0+\frac{3\pi}{2}\neq 0$ So C is false.

Option D: $xy''-4y'=0$

Compute: $xy''-4y'=x(-3x)-4\left(-\frac{3x^2}{2}\right)=-3x^2+6x^2=3x^2\neq 0$ So D is false.

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5. Final answer

The correct option is: $\boxed{\text{B}}$

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6. Comparison with stored answer

Stored correct answer: B

My derived answer: B

So they agree.