(α) We have
$$\begin{align*} x^3 - 3x \ge -2 &\Leftrightarrow x^3 - 3x + 2 \ge 0 \\ &\Leftrightarrow x^3 - x - 2x + 2 \ge 0 \\ &\Leftrightarrow x(x-1)(x+1) - 2(x-1) \ge 0 \\ &\Leftrightarrow (x-1)(x^2 + x - 2) \ge 0 \\ &\Leftrightarrow (x+2)(x-1)^2 \ge 0 \end{align*}$$
which is valid, because $x > 0$.

(β) The inequality is equivalent to
$$\frac{y}{z} \left( x^2 + \frac{2}{x} \right) + \frac{z}{x} \left( y^2 + \frac{2}{y} \right) + \frac{x}{y} \left( z^2 + \frac{2}{z} \right) \ge 9. \quad (2)$$
From question (α), by dividing both parts by $x$ we obtain $x^2 + \frac{2}{x} \ge 3$.
Therefore it is enough to prove:
$$3 \left( \frac{y}{z} + \frac{z}{x} + \frac{x}{y} \right) \ge 9 \Leftrightarrow \frac{y}{z} + \frac{z}{x} + \frac{x}{y} \ge 3. \quad (3)$$
The latter is valid, as we can easily see, by applying the inequality of arithmetic and geometric mean as follows:
$$\frac{y}{z} + \frac{z}{x} + \frac{x}{y} \ge 3 \sqrt{\frac{yzx}{zxy}} = 3 \quad (4)$$
The equality holds if and only if $x = y = z = 1$.