= Solution
Two nonzero <fractional ideals> $I,J$ of $\mathcal O_K$ are in the same ideal class when
$$
I=(\alpha)J
$$
for some $\alpha\in K^\times$, equivalently when $IJ^{-1}$ is principal. Multiplication is well defined on classes:
$$
[I][J]=[IJ].
$$
The identity is $[\mathcal O_K]$, the inverse of $[I]$ is $[I^{-1}]$, and commutativity comes from ideal multiplication. Nonzero fractional ideals are invertible because $\mathcal O_K$ is a Dedekind domain. Thus these classes form the abelian <ideal class group> $\operatorname{Cl}(K)$.
For finiteness, begin with a nonzero integral ideal $I$. By hypothesis choose $0\ne\alpha\in I$ with
$$
|N_{K/\mathbb Q}(\alpha)|\leq c_KN(I).
$$
Since $(\alpha)\subseteq I$, the ideal
$$
J=(\alpha)I^{-1}
$$
is integral, represents $[I]^{-1}$, and satisfies
$$
N(J)
=\frac{|N_{K/\mathbb Q}(\alpha)|}{N(I)}
\leq c_K.
$$
Hence every class has an integral representative of bounded norm. There are only finitely many integral ideals of bounded norm, so $\operatorname{Cl}(K)$ is finite. This is the <bounded-norm ideal representatives> argument.
Now take $K=\mathbb Q(\sqrt{-33})$. Since $-33\equiv3\pmod4$,
$$
\mathcal O_K=\mathbb Z[\sqrt{-33}],
\qquad
d_K=-132.
$$
The imaginary-quadratic <Minkowski bound for ideal classes> is
$$
\frac2\pi\sqrt{|d_K|}
=\frac2\pi\sqrt{132}<8.
$$
Every class is therefore represented by an integral ideal of norm at most seven.
The ramified prime ideals above two and three are
$$
\mathfrak p_2=(2,1+\sqrt{-33}),
\qquad
\mathfrak p_3=(3,\sqrt{-33}),
$$
with
$$
\mathfrak p_2^2=(2),
\qquad
\mathfrak p_3^2=(3).
$$
Neither is principal, since the norm equation
$$
a^2+33b^2=2\quad\text{or}\quad3
$$
has no integer solution. Their product is also nonprincipal, since $a^2+33b^2=6$ has no solution. Thus
$$
1,\quad[\mathfrak p_2],\quad[\mathfrak p_3],
\quad[\mathfrak p_2\mathfrak p_3]
$$
are four distinct classes, all of order at most two.
The prime five is inert, so it contributes no ideal of norm five. The prime seven splits as
$$
\mathfrak q_\pm=(7,\sqrt{-33}\mp3).
$$
Since
$$
(3+\sqrt{-33})
=\mathfrak p_2\mathfrak p_3\mathfrak q_-,
$$
comparison of norms, all equal to $42$, confirms the factorization and gives
$$
[\mathfrak q_-]=[\mathfrak p_2\mathfrak p_3].
$$
The conjugate prime has the inverse class, which is the same because this class has order two. Ideals of norms four and six yield respectively the principal class and $[\mathfrak p_2\mathfrak p_3]$. The Minkowski bound now shows that the four displayed classes exhaust the group. Therefore
$$
\boxed{
\operatorname{Cl}(\mathbb Q(\sqrt{-33}))
\cong C_2\times C_2
}.
$$
This is the <Ideal class group of Q of square root of minus thirty-three>.
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