Aurick throws $2$ fair $6$-sided dice labeled with the integers from $1$ through $6$. What is the probability that the sum of the rolls is a multiple of $3$?
1. **Define the problem and the generating function:**
   Aurick throws two fair 6-sided dice, and we need to find the probability that the sum of the rolls is a multiple of 3. The generating function for the sum of the dice rolls is:
   \[
   P(x) = (x + x^2 + x^3 + x^4 + x^5 + x^6)^2
   \]

2. **Roots of Unity Filter:**
   We use the Roots of Unity Filter to find the number of ways the sum can be a multiple of 3. Let the 3rd roots of unity be \(1, \omega, \omega^2\). The number of ways that the rolls can sum to a multiple of 3 is given by:
   \[
   \frac{P(1) + P(\omega) + P(\omega^2)}{3}
   \]

3. **Compute \(P(1)\):**
   \[
   P(1) = (1 + 1^2 + 1^3 + 1^4 + 1^5 + 1^6)^2 = (1 + 1 + 1 + 1 + 1 + 1)^2 = 6^2 = 36
   \]

4. **Compute \(P(\omega)\):**
   \[
   P(\omega) = (\omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6)^2
   \]
   Since \(\omega^3 = 1\), we have:
   \[
   \omega^3 = 1, \quad \omega^4 = \omega, \quad \omega^5 = \omega^2, \quad \omega^6 = 1
   \]
   Therefore:
   \[
   P(\omega) = (\omega + \omega^2 + 1 + \omega + \omega^2 + 1)^2 = (2\omega + 2\omega^2 + 2)^2
   \]
   The sum of the 3rd roots of unity is zero:
   \[
   \omega + \omega^2 + 1 = 0 \implies 2\omega + 2\omega^2 + 2 = 0 \implies P(\omega) = 0^2 = 0
   \]

5. **Compute \(P(\omega^2)\):**
   Similarly, we have:
   \[
   P(\omega^2) = (\omega^2 + (\omega^2)^2 + (\omega^2)^3 + (\omega^2)^4 + (\omega^2)^5 + (\omega^2)^6)^2
   \]
   Using \(\omega^3 = 1\), we get:
   \[
   (\omega^2 + \omega + 1 + \omega^2 + \omega + 1)^2 = (2\omega^2 + 2\omega + 2)^2
   \]
   Again, the sum of the 3rd roots of unity is zero:
   \[
   \omega + \omega^2 + 1 = 0 \implies 2\omega^2 + 2\omega + 2 = 0 \implies P(\omega^2) = 0^2 = 0
   \]

6. **Combine the results:**
   \[
   \frac{P(1) + P(\omega) + P(\omega^2)}{3} = \frac{36 + 0 + 0}{3} = 12
   \]

7. **Calculate the probability:**
   The total number of possible outcomes when rolling two dice is \(6^2 = 36\). Therefore, the probability that the sum of the rolls is a multiple of 3 is:
   \[
   \frac{12}{36} = \boxed{\frac{1}{3}}
   \]