If $y = e^{acos^{-1}x}, -1 < x < 1$, then $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx}$ is equal to

Given $y = e^{a\cos^{-1}x}$

Using the chain rule, and recalling that $\dfrac{d}{dx}(\cos^{-1}x) = \dfrac{-1}{\sqrt{1-x^2}}$:

$\dfrac{dy}{dx} = e^{a\cos^{-1}x} \cdot a \cdot \left(\dfrac{-1}{\sqrt{1-x^2}}\right) = \dfrac{-a \cdot y}{\sqrt{1-x^2}}$

Multiplying both sides by $\sqrt{1-x^2}$:

$\sqrt{1-x^2}\;\dfrac{dy}{dx} = -ay$

Squaring both sides:

$(1-x^2)\left(\dfrac{dy}{dx}\right)^2 = a^2y^2 \quad \cdots (i)$

Differentiating equation $(i)$ with respect to $x$:

$(-2x)\left(\dfrac{dy}{dx}\right)^2 + (1-x^2)\cdot 2\dfrac{dy}{dx}\cdot\dfrac{d^2y}{dx^2} = 2a^2 y\cdot\dfrac{dy}{dx}$

Since $y = e^{a\cos^{-1}x} > 0$ always, $\dfrac{dy}{dx} \neq 0$, so dividing both sides by $2\dfrac{dy}{dx}$:

$-x\dfrac{dy}{dx} + (1-x^2)\dfrac{d^2y}{dx^2} = a^2 y$

$(1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} = a^2 y$