= Solution
The polynomial is <Eisenstein> at $3$, so a root $\alpha$ generates a totally ramified cubic field $F=\mathbb Q_3(\alpha)$ and is a <uniformizer> of $F$. Its cubic discriminant is
$$
\Delta=-4(-3)^3-27\cdot3^2=-135=-5\cdot3^3.
$$
Its $3$-adic valuation is $3$, so it is not a square. The <cubic discriminant criterion> therefore gives
$$
\boxed{\operatorname{Gal}(M/\mathbb Q_3)\cong S_3,\qquad[M:\mathbb Q_3]=6.}
$$
The unique quadratic subfield is generated by $\sqrt\Delta$. Since the permitted square root $\sqrt{-5}$ is in $\mathbb Q_3$, this subfield is
$$
E=\mathbb Q_3(\sqrt3).
$$
It is totally ramified of degree two, by the <Eisenstein polynomial> $T^2-3$. The ramification index of $M/\mathbb Q_3$ is divisible by both the index $3$ of $F$ and the index $2$ of $E$. Since the whole degree is $6$, it follows that
$$
\boxed{e(M/\mathbb Q_3)=6,\qquad k_M=\mathbb F_3.}
$$
In particular $G_0=G$.
We use the following standard facts about totally ramified extensions: every uniformizer generates the valuation ring over the base valuation ring, and for a totally ramified Galois extension
$$
G_i=\{\sigma:v_M(\sigma(\pi)-\pi)\geq i+1\}\quad(i\geq0).
$$
Thus we can use the <uniformizer criterion for lower ramification groups>. Put $s=\sqrt3$. Since $v_M(\alpha)=2$ and $v_M(s)=3$,
$$
\pi=\frac{\alpha^2}{s},\qquad v_M(\pi)=4-3=1
$$
is a uniformizer of $M$.
Let $\alpha_1,\alpha_2,\alpha_3$ be the three roots. The <Galois group> acts transitively on their unordered pairs and preserves the valuation, so all differences between distinct roots have the same valuation $d$. The discriminant identity gives
$$
18=v_M(\Delta)=2\sum_{i<j}v_M(\alpha_i-\alpha_j)=6d,
$$
hence $d=3$. Thus, for any $\sigma$, either $\sigma(\alpha)=\alpha$ or $\sigma(\alpha)-\alpha$ has valuation $3$.
A nontrivial three-cycle fixes $s$, since it fixes the quadratic subfield, and sends $\alpha$ to a different root. Also $v_M(\sigma(\alpha)+\alpha)=2$: write the sum as $2\alpha+(\sigma(\alpha)-\alpha)$, whose two terms have unequal valuations $2,3$. Therefore
$$
v_M(\sigma(\pi)-\pi)=3+2-3=2
\quad\text{for either nontrivial three-cycle}.
$$
A transposition changes the sign of $s$. Consequently
$$
\sigma(\pi)-\pi=-\frac{\sigma(\alpha)^2+\alpha^2}{s}.
$$
The numerator has valuation $4$: it is $2\alpha^2$ plus terms of strictly higher valuation if the root changes, and exactly $2\alpha^2$ if the root is fixed. Thus every transposition has $v_M(\sigma(\pi)-\pi)=1$. We conclude that the <ramification groups of the splitting field of T3 minus 3T plus 3 over Q3> are
$$
\boxed{G_0=S_3,\qquad G_1=A_3\cong C_3,\qquad G_i=\{1\}\quad(i\geq2).}
$$
As an independent check, the cubic field has $\mathcal O_F=\mathbb Z_3[\alpha]$ and different exponent $v_F(g'(\alpha))=v_F(3(\alpha^2-1))=3$. The extension $M/F$ is tamely totally ramified of degree two, so its different exponent is $1$. The <different in a tower> gives $d(M/\mathbb Q_3)=1+2\cdot3=7$, agreeing with the ramification sum $(6-1)+(3-1)=7$.
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