Answer
$$a=-3,\ \ a=4 $$
Work Step by Step
Given $$y^{\prime \prime}-12 x^{-2} y=0$$
Since $ y=x^a$ is a solution, then
\begin{align*}
y'&= ax^{a-1}\\
y''&= a(a-1) x^{a-2}
\end{align*}
Hence
\begin{align*}
y^{\prime \prime}-12 x^{-2} y&=0\\
a(a-1) x^{a-2}-12x^{-2} x^a&=0\\
[a(a-1) -12 ]x^{a-2}&=0
\end{align*}
Then $$a^2-a-12=0\to (a-4)(a+3)=0$$
Hence $$a=-3,\ \ a=4 $$
