= Solution
Let $a=\sqrt[3]{3}$ and let $\zeta$ be a primitive cube root of unity. The <splitting field> is $L=\mathbb Q_3(a,\zeta)$. The <polynomial> $X^3-3$ is an <Eisenstein polynomial>, so $[\mathbb Q_3(a):\mathbb Q_3]=3$. Meanwhile $(1+2\zeta)^2=-3$ shows that $\mathbb Q_3(\zeta)$ is quadratic: $-3$ is not a square because its <valuation> is odd. A quadratic field cannot lie in the degree-three field. Consequently $[L:\mathbb Q_3]=6$, and the action on the three roots identifies its <Galois group> with the <symmetric group> $S_3$.
A useful <uniformizer> is
$$
\pi=\frac{\zeta-1}{a}.
$$
Since $(\zeta-1)^3=-3\zeta(\zeta-1)$, we obtain
$$
\pi^3=-\zeta(\zeta-1)=1+2\zeta,\qquad\pi^6=-3.
$$
Thus $\pi$ is a root of the <Eisenstein polynomial> $X^6+3$. Its degree is six, so $L=\mathbb Q_3(\pi)$, the extension is totally ramified, and $\pi$ is a <uniformizer>. With $v_L(\pi)=1$,
$$
v_L(3)=6,\qquad v_L(a)=2,\qquad v_L(\zeta-1)=3.
$$
This is the <Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3>.
Let $\tau(a)=\zeta a$, $\tau(\zeta)=\zeta$, and let $s(a)=a$, $s(\zeta)=\zeta^{-1}$. Then $\tau$ has order three, $s$ has order two, and $s\tau s=\tau^{-1}$. Their actions on the <uniformizer> are
$$
\tau(\pi)=\zeta^{-1}\pi,\qquad s(\pi)=-\zeta^{-1}\pi.
$$
The two nonidentity elements of $\langle\tau\rangle$ therefore satisfy
$$
v_L(\tau(\pi)-\pi)=v_L(\tau^2(\pi)-\pi)=1+v_L(\zeta-1)=4.
$$
Each transposition sends $\pi$ to $-\zeta^j\pi$ for some $j$, so its displacement has <valuation> one: $1+\zeta^j$ is a unit, reducing to $2$ modulo the <maximal ideal>. The <uniformizer criterion for lower ramification groups> now gives
$$
\boxed{G_{-1}=G_0=S_3,\qquad G_1=G_2=G_3=\langle\tau\rangle\cong C_3,\qquad G_i=1\ (i\geq4).}
$$
For real indices, this is $S_3$ on $[-1,0]$, $C_3$ on $(0,3]$, and $1$ for $t>3$. In particular the lower breaks are $0$ and $3$.
On $(0,3]$ the index $[G_0:G_t]$ is two; beyond three it is six. Thus the <Herbrand function> is
$$
\varphi(t)=\begin{cases}t,&-1\leq t\leq0,\\t/2,&0\leq t\leq3,\\3/2+(t-3)/6,&t\geq3.\end{cases}
$$
The positive lower break $3$ becomes the upper break $3/2$. Therefore
$$
\boxed{G^u=\begin{cases}S_3,&-1\leq u\leq0,\\C_3,&0<u\leq3/2,\\1,&u>3/2.\end{cases}}
$$
The fractional upper break is allowed because the full extension is nonabelian. As a check, the <different exponent from ramification groups> is $5+3\cdot2=11$. The derivative of the <minimal polynomial> $X^6+3$ gives the same result: $v_L(6\pi^5)=6+5=11$. This is the $p=3$ case of <ramification groups of the splitting field of Xp minus p over the p-adic numbers>.
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