It is given that, $y=e^{a \cos ^{-1} x}$

Taking logarithm on both the sides, we obtain

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

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

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \frac{d y}{d x}=a \times \frac{-1}{\sqrt{1-x^{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-a y}{\sqrt{1-x^{2}}}$

By squaring both the sides, we obtain

$\left(\frac{d y}{d x}\right)^{2}=\frac{a^{2} y^{2}}{1-x^{2}}$

$\Rightarrow\left(1-x^{2}\right)\left(\frac{d y}{d x}\right)^{2}=a^{2} y^{2}$

$\left(1-x^{2}\right)\left(\frac{d y}{d x}\right)^{2}=a^{2} y^{2}$

Again differentiating both sides with respect to $x$, we obtain

$\left(\frac{d y}{d x}\right)^{2} \frac{d}{d x}\left(1-x^{2}\right)+\left(1-x^{2}\right) \times \frac{d}{d x}\left[\left(\frac{d y}{d x}\right)^{2}\right]=a^{2} \frac{d}{d x}\left(y^{2}\right)$

$\Rightarrow\left(\frac{d y}{d x}\right)^{2}(-2 x)+\left(1-x^{2}\right) \times 2 \frac{d y}{d x} \cdot \frac{d^{2} y}{d x^{2}}=a^{2} \cdot 2 y \cdot \frac{d y}{d x}$

$\Rightarrow\left(\frac{d y}{d x}\right)^{2}(-2 x)+\left(1-x^{2}\right) \times 2 \frac{d y}{d x} \cdot \frac{d^{2} y}{d x^{2}}=a^{2} \cdot 2 y \cdot \frac{d y}{d x}$

$\Rightarrow-x \frac{d y}{d x}+\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} \cdot y$   $\left[\frac{d y}{d x} \neq 0\right]$

$\Rightarrow\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$

Hence, proved.