1. **Lemma: If $n$ is a perfect number, then $mn$, where $m > 1$, is not a perfect number.

Proof of Lemma:**
Let $d_1, d_2, \ldots, d_k$ be the proper divisors of $n$ such that $d_1 + d_2 + \ldots + d_k = n$. Then, for $mn$, the proper divisors are $md_1, md_2, \ldots, md_k$ and possibly other divisors. Note that $md_i$ for $1 \le i \le k$ are distinct divisors of $mn$ and $1 < md_i < mn$. If $mn$ were a perfect number, we would have:
$$mn = 1 + md_1 + md_2 + \ldots + md_k$$
However, this implies:
$$mn \ge 1 + mn$$
which is a contradiction. Hence, $mn$ is not a perfect number. $\blacksquare$

2. Observation:
The number $6$ is a perfect number because $6 = 1 + 2 + 3$.

3. Assumption:
Assume there is a perfect number $n$ greater than $6$ that is divisible by $3$ but not by $9$. Let $n = 3k$.

4. Sum of Divisors Function:
Let $f(m)$ be the sum of the divisors of a number $m$. Note that $f$ is multiplicative, so:
$$f(n) = f(3k) = f(3)f(k)$$
Since the sum of the divisors of $3$ is $1 + 3 = 4$, we have:
$$f(n) = 4f(k)$$

5. Perfect Number Condition:
For $n$ to be a perfect number, the sum of its proper divisors must equal $n$. Therefore:
$$n = f(n) - n$$
Substituting $f(n) = 4f(k)$ into the equation, we get:
$$n = 4f(k) - n$$
Solving for $n$, we have:
$$2n = 4f(k) \implies n = 2f(k)$$
This implies that $2$ divides $n$, and since $n = 3k$, $6$ must divide $n$.

6. Contradiction:
By the lemma, $n$ cannot be a perfect number if it is of the form $3k$ and not divisible by $9$. This leads to a contradiction.

Therefore, $n$ must be divisible by $9$.

$\blacksquare$