1. Given relation

   $\ln y = 3\sin^{-1}x$

   So,

   $y = e^{3\sin^{-1}x}$

2. Differentiate once

   Differentiating $\ln y = 3\sin^{-1}x$ with respect to $x$:

   $\frac{y'}{y} = \frac{3}{\sqrt{1-x^2}}$

   Hence,

   $y' = \frac{3y}{\sqrt{1-x^2}}$

3. Differentiate again

   Differentiate

   $y' = 3y(1-x^2)^{-1/2}$

   Using product rule:

   $y'' = 3y'(1-x^2)^{-1/2} + 3y\cdot \frac{d}{dx}(1-x^2)^{-1/2}$

   Now,

   $\frac{d}{dx}(1-x^2)^{-1/2} = \frac{x}{(1-x^2)^{3/2}}$

   Therefore,

   $y'' = \frac{3y'}{\sqrt{1-x^2}} + \frac{3xy}{(1-x^2)^{3/2}}$

   Substitute $y' = \frac{3y}{\sqrt{1-x^2}}$:

   $y'' = \frac{3}{\sqrt{1-x^2}}\cdot \frac{3y}{\sqrt{1-x^2}} + \frac{3xy}{(1-x^2)^{3/2}}$

   $y'' = \frac{9y}{1-x^2} + \frac{3xy}{(1-x^2)^{3/2}}$

4. Compute $(1-x^2)y'' - xy'$

   Multiply $y''$ by $(1-x^2)$:

   $(1-x^2)y'' = 9y + \frac{3xy}{\sqrt{1-x^2}}$

   Also,

   $xy' = x\cdot \frac{3y}{\sqrt{1-x^2}} = \frac{3xy}{\sqrt{1-x^2}}$

   Hence,

   $(1-x^2)y'' - xy' = 9y$

5. Evaluate at $x=\frac12$

   First,

   $\sin^{-1}\left(\frac12\right)=\frac\pi6$

   So,

   $y = e^{3\cdot \pi/6} = e^{\pi/2}$

   Therefore,

   $(1-x^2)y'' - xy' = 9y = 9e^{\pi/2}$

6. Option check

   • A: $9e^{\pi/2}$ ✅
   • B: $9e^{\pi/6}$ ❌
   • C: $3e^{\pi/2}$ ❌
   • D: $3e^{\pi/6}$ ❌

Therefore, the correct answer is A.