Answer
If $ x \neq 0$, then $y = \frac{1}{x}$ satisfies $x^3y'' + x^2y' - xy =0$.
Work Step by Step
If $y = \frac{1}{x} = x^{-1}$, then
\begin{align}
y' &= (-1)x^{-1-1} \\
&= -x^{-2} \\
y'' &= (-2)(-x^{-2-1}) \\
&= 2x^{-3} \\
\end{align}
Substitution $y, y'$, and $y''$ into the differential equation, we obtain
$x^3(\frac{2}{x^3}) + x^2(-\frac{1}{x^2}) - x(\frac{1}{x}) = 2 -1 - 1 = 0 $