A useful way to solve these sorts of problems is as follows:
Write $\alpha = 1 + x \lambda,$ for some $x \in \mathbb Z[\omega]$. (This is possible by the assumption that $\alpha \equiv 1 \bmod \lambda$.)
Cubing both sides gives
$$\alpha^3 = 1 + 3 x \lambda + 3 x^2 \lambda^2 + x^3 \lambda^3
= 1 +(- \omega^2 x + x^3)\lambda^3 - \omega^2 x^2 \lambda^4$$
(the last equality following from $3 = - \omega^2 \lambda^2$).
The point is that I have written $\alpha^3$ as a sum of multiples of increasing powers of $\lambda$.
What you want to show is that $\alpha^3 \equiv 1 \bmod 9$, or equivalently $\alpha^3 \equiv 1 \bmod \lambda^4$, and so looking at the preceding expression,
we see that this is the same as proving
$-\omega^2 x + x^3 \equiv 0 \bmod \lambda,$ for all $x$ (because any choice of $x$ in the original formula for $\alpha$ gives a choice of $\alpha$ which is $1 \bmod \lambda$).
Since $\omega \equiv 1 \bmod \lambda,$ we see that the congruence we have to prove can be simplified to
$$x^3 - x \equiv 0 \bmod \lambda$$
for all $x \in \mathbb Z[\omega]$.
Why is this true? Because $Z[\omega]/ \lambda Z[\omega]$ is a field of order $3$ (i.e. a copy of $\mathbb F_3$), and in that field every element is its own $3$rd power. (In general in a field of order $p$, we always have $x^p = x$.) So we're done.
The nice thing about this approach (expanding in powers of $\lambda$) is that you can follow your nose in the proof: you turn the congruence you have to prove into a simpler congruence. (Our first congruence was something subtle about $\alpha$ that are $1 \bmod \lambda$, but we converted it into a congruence that had to be true for all $x$, and if a congruence holds for all $x$, it will normally have to hold for some fairly obvious reason.)
One last remark: in congruences involving powers, the binomial theorem
is very often helpful (as it was here).