Note that the ring of integers of $\mathbb{Q}(\sqrt[3]{3})$ is $\mathbb{Z}[\sqrt[3]{3}]$ with basis $1,\sqrt[3]{3},\sqrt[3]{9}$ over $\mathbb{Z}$. You want to show that the norm, defined by
\begin{eqnarray*}
N(a+b\sqrt[3]{3} + c\sqrt[3]{9}) & = & (a+b\sqrt[3]{3} + c\sqrt[3]{9})(a+\zeta_3b\sqrt[3]{3} + \zeta_3^2c\sqrt[3]{9})(a+\zeta_3^2b\sqrt[3]{3} + \zeta_3c\sqrt[3]{9})\\
& = & a^3 + 3b^3 + 9c^3 - 9abc
\end{eqnarray*}
gives in fact a Euclidean norm upon taking absolute values, where $\zeta_3$ is a fixed primitive cube root of unity. Now, the first (and maybe the main) thing you might wonder about is where I pulled this norm out. For that you have to know a little bit of Galois theory. The basic idea is that from the perspective of $\mathbb{Q}$, the elements $\sqrt[3]{3}$ and $\zeta_3\sqrt[3]{3}$ are indistinguishable: they are both just some roots of the polynomial $x^3-3$ and can be thought of as mirror images of each other. Essentially, the factors that I am multiplying are like "mirror images" of my given element of $\mathbb{Z}[\sqrt[3]{3}]$.
Now, the strategy is exactly the same as in the case of some quadratic rings like $\mathbb{Z}[\sqrt{2}]$: given $\alpha = u+v\sqrt[3]{3} + w\sqrt[3]{9}$ and
$\beta = f+g\sqrt[3]{3} + h\sqrt[3]{9}\in \mathbb{Z}[\sqrt[3]{3}]$, you want to show that there exist $p$ and $r$ in that ring such that
\begin{eqnarray*}\alpha = p\beta + r,
\end{eqnarray*}
where either $r=0$ or $N(r)