Substitution is blind - 1

$\begin{aligned} y & = \sqrt[3] 3 + \frac 1{\sqrt[3] 3} & \small \color{#3D99F6} \text{Cubing both sides} \\ y^3 & = 3 + 3\sqrt[3] 3 + \frac 3{\sqrt[3]3} + \frac 13 \\ y^3 & = 3 + 3y + \frac 13 & \small \color{#3D99F6} \text{Multiplying both sides by }3 \\ 3y^3 & = 9 + 9y + 1 & \small \color{#3D99F6} \text{Rearranging} \end{aligned}$

$\begin{aligned} \implies 3y^3 - 9y & = \boxed{10} \end{aligned}$

Use the relation $(a+b)^3=a^3+b^3+3ab(a+b)$ to get $y^3=\left(\sqrt[3]3+\dfrac{1}{\sqrt[3]3}\right) ^3=3+\dfrac{1}{3}+3y=\dfrac{10}{3}+3y\implies 3y^3-9y=\boxed {10}$ .