If $$\log_e y = 3\sin^{-1} x$$, then $$(1 - x^2)y'' - xy'$$ at $$x = \frac{1}{2}$$ is equal to

$$\ln y = 3\sin^{-1}x$$. Differentiating: $$\frac{y'}{y} = \frac{3}{\sqrt{1-x^2}}$$, so $$y' = \frac{3y}{\sqrt{1-x^2}}$$.

$$(1-x^2)(y')^2 = 9y^2$$. Differentiating: $$(1-x^2)2y'y'' - 2x(y')^2 = 18yy'$$.

Dividing by $$2y'$$: $$(1-x^2)y'' - xy' = 9y$$.

At $$x = 1/2$$: $$\ln y = 3\sin^{-1}(1/2) = 3\pi/6 = \pi/2$$, so $$y = e^{\pi/2}$$.

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

The correct answer is Option 2: $$9e^{\pi/2}$$.