1. Define the problem and variables:
We need to find all 6-digit natural numbers consisting only of the digits $1, 2,$ and $3$, where the digit $3$ occurs exactly twice, and the number is divisible by $9$.

2. Set up the variables:
Let $N$ be such a number. Let $x$ be the number of $1$'s and $y$ be the number of $2$'s in $N$. Since $N$ is a 6-digit number and $3$ occurs exactly twice, we have:
$$x + y + 2 = 6 \implies x + y = 4$$

3. Condition for divisibility by 9:
A number is divisible by $9$ if the sum of its digits is divisible by $9$. The sum of the digits of $N$ can be expressed as:
$$1 \cdot x + 2 \cdot y + 3 \cdot 2 = x + 2y + 6$$
For $N$ to be divisible by $9$, we need:
$$x + 2y + 6 \equiv 0 \pmod{9}$$

4. **Substitute $x + y = 4$ into the divisibility condition:**
From $x + y = 4$, we can express $y$ as $y = 4 - x$. Substitute this into the divisibility condition:
$$x + 2(4 - x) + 6 \equiv 0 \pmod{9}$$
Simplify the equation:
$$x + 8 - 2x + 6 \equiv 0 \pmod{9}$$
$$-x + 14 \equiv 0 \pmod{9}$$
$$-x \equiv -14 \pmod{9}$$
$$-x \equiv -5 \pmod{9}$$
$$x \equiv 5 \pmod{9}$$

5. **Check the possible values of $x$:**
Since $x$ must be a positive integer and $x \leq 4$ (because $x + y = 4$), we check if $x = 5$ is possible:
$$x = 5$$
However, $x = 5$ is not possible because $x$ must be less than or equal to $4$.

6. Conclusion:
Since there is no valid $x$ that satisfies the conditions, no such 6-digit number exists.

$\blacksquare$