Prove $u$ is a unit if and only if $N(u) = 1$? The norm $N:\mathbb{Z}[\sqrt[3]{2}] \rightarrow \mathbb{N}$ defined by $N(a+b\sqrt[3]{2} + c\sqrt[3]{4}) = |a^3 + 2b^3 + 4c^3 - 6abc|$ is multiplicative. (Already proven). Show that $\alpha \in \mathbb{Z}[\sqrt[3]{2}]$ is a unit if and only if $N(\alpha) = 1$.
So far proved it one direction, but am struggling with the other.
Proof $\rightarrow$
Let $\alpha$ be a unit, so by defintion $\exists \beta$ such that $\alpha\beta = 1$. We can then use the norm, knowing that $N(\alpha\beta) = |(1)^3 + 2(0)^3 + 4(0)^3 - 6(1)(0)(0)| = 1$. Since the norm is multiplicative, $N(\alpha\beta) = N(\alpha)N(\beta) = 1$, and since the results of the norms have to be positive integers, we know $N(\alpha)=1$.
Proof $\leftarrow$ Let $N(\alpha)$ = 1. Then for $\alpha = a+b\sqrt[3]{2} + c\sqrt[3]{4}$, $|a^3 + 2b^3 + 4c^3 - 6abc|=1$
Any help on where to go?

Hint: if you embed $\mathbb{Z}[\sqrt[3]{2}]$ into $\mathbb{Z}[\sqrt[3]{2}, \omega]$ where $\omega := e^{2 \pi i / 3}$, then
$$N(a + b \sqrt[3]{2} + c \sqrt[3]{4}) = \left| (a + b \sqrt[3]{2} + c \sqrt[3]{4}) (a + b \omega \sqrt[3]{2} + c \omega^2 \sqrt[3]{4}) (a + b \omega^2 \sqrt[3]{2} + c \omega \sqrt[3]{4}) \right|.$$
(This expression is heavily inspired by the definition of norm in Galois theory, of which your definition is close to being a special case.)
Now, multiply the last two factors on the right hand side, and you should get an expression for $\pm$ of the inverse of $a + b \sqrt[3]{2} + c \sqrt[3]{4}$ which lies in $\mathbb{Z}[\sqrt[3]{2}]$.  (With the sign being determined by the sign of $a^3 + 2 b^3 + 4 c^3 - 6abc$.)
