These inequalities are surely true when $x = 0$, so assume $x > 0$ and let $y = x + \frac{1}{x}$, so that $y \ge 2$, and
$$y^3 - 3y = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} - 3\left(x + \frac{1}{x}\right) = x^3 + \frac{1}{x^3} = \frac{x^6 + 1}{x^3}.$$
Also
$$\frac{3x^2 - 2x + 3}{4} = \frac{x}{4} \left( 3x - 2 + \frac{3}{x} \right) = x \cdot \frac{3y - 2}{4},$$
and
$$2x^2 - 3x + 2 = x(2y - 3).$$
Hence, we are required to prove that
$$\frac{3y - 2}{4} \le \sqrt[3]{\frac{y^3 - 3y}{2}} \le 2y - 3, \quad \text{for all } y \ge 2.$$
Consider the rightmost inequality. It holds iff
$$\begin{aligned} y^3 - 3y &\le 2(2y - 3)^3 = 16y^3 - 72y^2 + 108y - 54, \quad \text{i.e.} \\ 0 &\le 15y^3 - 72y^2 + 111y - 54 = 3(y - 1)(y - 2)(5y - 9), \end{aligned}$$
which is true since each factor is non-negative. Clearly, this inequality is strict unless $y = 2$, i.e. $x = 1$.
We treat the leftmost one similarly, which is equivalent to the claim that $(3y - 2)^3 \le 32(y^3 - 3y)$ for all $y \ge 2$. Expanding and simplifying this becomes
$$0 \le 5y^3 + 54y^2 - 132y + 8 = (y - 2)(5y^2 + 62y + 2(y - 2)),$$
which is true, with equality iff $y = 2$. The result follows.