Let $D = \Bbb Z[\sqrt{-6}]$ and $P = (2)$.
I'm trying to show that $D/P$ is an integral domain (at least I think this might be the case). Clearly if $\alpha\beta \in P$ then $\alpha\beta = 2r$ for some $r \in D$. If $\alpha = a_1 + b_1\sqrt{-6}$ and $\beta = a_2 + b_2\sqrt{-6}$ then we require that $a_1a_2 - 6b_1b_2 = 2a_3$ and $a_1b_2 + a_2b_1 = 2b_3$ where $r = a_3 + b_3\sqrt{-6}$.
In the first condition, we have what we want; the $6b_1b_2$ term is always divisible by two so either $2 \mid a_1$ or $2 \mid a_2$, which is what we want. I think the second condition gives too much flexibility, though. We need $2 \mid a_1$ or $2 \mid a_2$ but if $2\mid a_1$ then we could have $2\mid b_2$ and so $\alpha \notin P$ and $\beta \notin P$ meaning that $(2)$ is not a prime ideal of $D$.
Is this reasoning correct? (I realise that my "thinking $D/P$ is an integral domain statement is now false by my own reasoning, but I'll leave it in for laughs)
$$\mathbf Z[\sqrt{-6}] \simeq \mathbf Z[X]/(X^2+6),$$ \begin{align}\text{so}\hspace8em\mathbf Z[\sqrt{-6}]/(2)&\simeq \mathbf Z[X]/(X^2+6)\Big/2 \cdot\mathbf Z[X]/(X^2+6)\\&\simeq \mathbf Z[X]/(X^2+6)\Big/(2,X^2+6)/(X^2+6) \\ &\simeq \mathbf Z/2\mathbf Z[X]\Big/(\bar 2,X^2+\bar6)= \mathbf Z/2\mathbf Z[X]/(X^2) \end{align} This shows $(2)$ is ramified in $\mathbf Z[\sqrt{-6}]$. For the same reason, $(3)$ is ramified.