Answer
$y(x)=c_1x^3+c_2x^{-3}$
Work Step by Step
Given $x^2y''-6y=0$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$$r(r-1)-6=0\\
r^2-r-6=0$$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=x^3\\
y_2(x)=x^{-3}$
so that the general solution is
$y(x)=c_1x^3+c_2x^{-3}$
