Solution: Let $y(x) = x \cdot v(x)$ be the solution of the given differential equation. We need to determine $v(x)$. Then we have \begin{align*} y'(x) & = xv'(x) + v(x) \\ y''(x) & = xv''(x) + 2v'(x) \\ y'''(x) & = xv'''(x) + 3v''(x). \end{align*} Therefore, the given differential equation reduces to \begin{align*} & x^3 \left( 3v'' + xv''' \right) - 3x^2 \left( 2v' + xv'' \right) \\ & \kern 2cm + (6x - x^3) \left( v + xv' \right) + \left( x^2 - 6 \right) xv = 0 \\ \implies & x^4 v''' + v''\left( 3x^3 - 3x^3 \right) + v'\left( -6x^2 + 6x^2 - x^4 \right) \\ & \kern 2cm + v\left( 6x - x^3 + x^3 - 6x \right) = 0 \\ \implies & x^4 \left( v''' - v' \right) = 0 . \end{align*} This implies $v'''(x) - v'(x) = 0$, and the solution to this differential equation is \[ v(x) = c_1 + c_2 e^x + c_3 e^{-x}. \] Therefore, the solution of the given differential equation will be \[ y(x) = c_1 x + c_2 x e^x + c_3 x e^{-x}. \]