Solution: The given function is \[ f(x) = \frac{x}{3} + \frac{3}{x} + 3, \ x \neq 0. \] Observe that \begin{align*} f'(x) = \frac{1}{3} - \frac{3}{x^2 }. \end{align*} Therefore, \begin{align*} f \text{ is strictly increasing if } f'(x) > 0 & \implies \frac{1}{3} - \frac{3}{x^2} >0 \\ & \implies \frac{x^2 - 9}{3x^2} > 0 \\ & \implies x^2 - 9 > 0 \\ & \implies (x-3)(x+3) > 0\\ & \implies x \in (-\infty ,-3) \cup (3, \infty ). \\[2ex] f \text{ is strictly decreasing if } f'(x) < 0 & \implies (x-3)(x+3) < 0\\ & \implies x\in (-3,3) \setminus \{ 0 \} \\ & \implies x \in (-3, 0) \cup (0,3). \end{align*} Therefore, $\alpha _1 = -3, \alpha _2 = 3, \alpha _3 = -3, \alpha _4 = 0$ and $\alpha _5 = 3$. Thus, \begin{align*} \sum_{i=1}^{5} \alpha _i^2 = 4 \times 3^2 = 36. \end{align*}