$\langle 3 \rangle$ in $\textbf{Z}[\omega]$ ramifies, not splits, right? Potentially dumb question here, please bear with me, I want to make sure I have not overlooked some important subtlety.
As you know, $$\omega = \frac{-1 + \sqrt{-3}}{2}$$ is a complex cubit root of unity, that is, $\omega^3 = 1$. This slightly obscures the fact that $1 + 2 \omega = \sqrt{-3}$, and since $(\sqrt{-3})^2 = -3$, it follows that $\langle 3 \rangle = \langle \sqrt{-3} \rangle^2$. Meaning that $\langle 3 \rangle$ ramifies, it doesn't split. This even though $$\left( \frac{3 - \sqrt{-3}}{2} \right) \left( \frac{3 + \sqrt{-3}}{2} \right) = 3.$$
That's because $$\left( \frac{-1 + \sqrt{-3}}{2} \right) \sqrt{-3} = \frac{-\sqrt{-3} - 3}{2},$$ or $\omega(1 + 2 \omega) = \omega + 2 \omega^2$. Since $\omega$ is a unit, the ideals $\langle 1 + 2 \omega \rangle$ and $\langle \omega + 2 \omega^2 \rangle$ are in fact the same.
Are these calculations correct? Have I drawn the right conclusion?

It seems fine, although I must admit I had trouble understanding what you want.
Anyway, first to conclude that $\langle 3 \rangle $ ramifies you need to prove that $\left \langle \sqrt{-3} \right \rangle$ is a prime ideal. This shouldn't be too hard to see, as it's norm is $3$ a prime number.
Then you can establish the following relation
$$\left \langle \sqrt{-3} \right \rangle = \left \langle \frac{3 - \sqrt{-3}}{2} \right \rangle = \left \langle \frac{3 + \sqrt{-3}}{2} \right \rangle$$
to conclude that the both ways lead to the same prime ideal factorization of $\langle 3 \rangle$. As you have already noticed we have that $\left \langle \sqrt{-3} \right \rangle = \left \langle \frac{3 + \sqrt{-3}}{2} \right \rangle$, as $\left( \frac{1 - \sqrt{-3}}{2} \right) \sqrt{-3} = \frac{3 + \sqrt{-3} }{2}$
Similarly $\left \langle \sqrt{-3} \right \rangle = \left \langle \frac{3 - \sqrt{-3}}{2} \right \rangle$, as $\left( \frac{-1 - \sqrt{-3}}{2} \right) \sqrt{-3} = \frac{3 - \sqrt{-3} }{2}$ and the left-most factor is $\omega^2$, another unit.
