I aborted trying to reduce that monster mod 6561 by hand, but I'll still go through with the algorithm by hand.
First, because everything that matters is mod powers of $3$, we'll do our arithmetic base 9. We wish to solve $2^x\equiv 2732_9\mod 10000_9$.
The first step of the algorithm is to split things up based on the prime factors of the order of the group we're working in. Here, that order is $n=\phi(3^8)=2\cdot 3^7$.
First, powers of $2$. Raise both the generator and the objective to the $\frac{n}{2}$ power; we wish to solve $\left(2^{3^7}\right)^x\equiv(2732_9)^{3^7}$ - or, simply, $(-1)^x\equiv-1$. (We only need to know what $k$ is mod $3$ to determine its $3^7$ power mod $3^8$). The solution is $x\equiv 1\mod 2$.
Next, powers of $3$ - the interesting part. Raise both generator $g$ and objective $h$ to the $\frac{n}{3^7}=2$ power: $2^2\equiv 4$ and
$$2732_9^2 = \begin{array}{r}5564\phantom{000}\\ 21645\phantom{00}\\ 8406\phantom{0}\\5564\\ \hline 7840234\end{array}\equiv 0234_9$$
Now we get down to the prime-power algorithm, solving $4^x\equiv 0234_9\mod 10000_9$ (in the order $3^7$ subgroup). First, we need to compute $\gamma \equiv 4^{3^6}$. $(1+3)^{3^6}=1+3\cdot 3^6+3^2\cdot \binom{3^6}{2}+3^3\cdot\binom{3^6}{3}+\cdots$ and all but the first two terms are divisible by $3^8$, so $\gamma = 4^{3^6}\equiv 3001_9\mod 10000_9$.
Now, the core loop. We run $k$ from $0$ to $6$, first computing $h_k=\left(g^{-x_k}h\right)^{3^{6-k}}$, then finding $d_k$ such that $\gamma^{d_k}=h_k$, then setting $x_{k+1}=x_k+3^k d_k$.
While the wiki page I copied the algorithm suggests the "baby-step giant-step" algorithm for finding $d_k$, we won't do that here - $d_k$ ranges from $0$ to $p-1$, and for $p=3$ we can find it by inspection.
$k=0$:
$x_0$ is initialized as zero, of course. We have $h_0\equiv (0234_9)^{3^6} \equiv 4+(3^2\cdot 23_9)^{3^6} \equiv 4^{3^6}+3^8\cdot\text{stuff}\equiv 3001_9$.
Then $3001_9^1\equiv 3001_9$ and $d_0=1$.
Finally, $x_1= 0 + 1\cdot 1 = 1$.
$k=1$:
$h_1\equiv (4^{-1}\cdot 0234_9)^{3^5}\equiv (6731_9)^{3^5}\equiv (1+3^3\cdot\text{stuff})^{3^5}\equiv 1$.
This makes $d_1=0$.
Finally, $x_2=1+3\cdot 0=1$.
$k=2$:
$h_2\equiv (4^{-1}\cdot 0234_9)^{3^4}\equiv (6731_9)^{3^4}\equiv (1+3^3\cdot(1+3\cdot \text{stuff}))^{3^4}\equiv 1 +3^7\cdot(1+3\cdot \text{stuff})\equiv 3001_9$.
This makes $d_2=1$.
Finally, $x_3=1+3^2\cdot 1=11_9$.
$k=3$:
$h_3\equiv (4^{-11_9}\cdot 0234_9)^{3^3}$. This takes some calculation, but $4^9\equiv 8531_9$ and $4^{-9}\equiv 4361_9$. Then $4361_9\cdot 6731_9\equiv 1\color{red}{2}01$. Now $h_3\equiv (1201_9)^{3^3}\equiv \left(1+3^4(2+3^2)\right)^{3^3}\equiv 1+2\cdot 3^7+\text{stuff}\equiv 6001_9$.
This makes $d_3=2$.
Finally, $x_4=11_9+3^3\cdot 2=71_9$.
$k=4$:
$h_4\equiv (4^{-71_9}\cdot 0234_9)^{3^2}\equiv \left(4^{-2\cdot 3^3}\cdot 1201_9\right)^{3^2}$. We have $4^{3^3}\equiv 1701_9$ and $4^{-3^3}\equiv 7201_9$; square and multiply by $1201_9$ for $(1200_9+7200_9+7200_9)+1\equiv 6601_9$. Now $h_4\equiv (6601_9)^{3^2}\equiv 6001_9$.
This makes $d_4=2$.
Finally, $x_5=71_9+3^4\cdot 2=271_9$.
$k=5$:
$h_5\equiv (4^{-271_9}\cdot 0234_9)^3\equiv \left(4^{-2\cdot 3^4}\cdot 6601_9\right)^3$. We have $4^{3^4}\equiv 1701_9^3\equiv 1+3\cdot 1701_9\equiv 5301_9$ and $4^{-3^4}\equiv 3601_9$; square and multiply by $6601_9$ for $5001_9$. Now $h_5\equiv (5001_9)^3\equiv 6001$.
This makes $d_5=2$.
Finally, $x_6=271_9+3^5\cdot 2=871_9$.
$k=6$:
$h_6\equiv (4^{-871_9}\cdot 0234_9)^1\equiv \left(4^{-2\cdot 3^5}\cdot 5001_9\right)^1$. We have $4^{-3^5}\equiv 2001$; squaring and multiplying gives $h_6\equiv 1$.
This makes $d_6=0$.
Finally, $x_7=871_9+3^6\cdot 0=871_9$.
That's our result: $x\equiv 871_9\mod 3^7$. Converting back to base ten, that's $x\equiv 8\cdot 81+7\cdot 9+1\equiv 712\mod 2187$.
The final step is to combine what we got for powers of 2 and powers of 3. We want something that's $1$ mod $2$ and $712$ mod $2187$; final answer $2899\mod 4374$.
My first-run answer doesn't match. Time to check it... Spreadsheet-aided checking found the error, marked in red (1101 by initial calculation, leading to $3025$ final answer). Recalculating from there forward... OK, got it now.
While the work involved in doing this by hand isn't too hard, it's so long that it's really easy to make an arithmetic mistake leading to a wrong answer. I wouldn't recommend it without at least machine-checking the various arithmetic pieces. No particularly sophisticated tools are needed for that, at least.
