Juan rolls a fair regular octahedral die marked with the numbers $ 1$ through $ 8$. Then Amal rolls a fair six-sided die. What is the probability that the product of the two rolls is a multiple of $ 3$?

$ \textbf{(A)}\ \frac{1}{12} \qquad
\textbf{(B)}\ \frac{1}{3} \qquad
\textbf{(C)}\ \frac{1}{2} \qquad
\textbf{(D)}\ \frac{7}{12} \qquad
\textbf{(E)}\ \frac{2}{3}$
To determine the probability that the product of the two rolls is a multiple of 3, we need to consider the possible outcomes for each die and the conditions under which their product is a multiple of 3.

1. **Identify the numbers on each die:**
   - The octahedral die has numbers: \(1, 2, 3, 4, 5, 6, 7, 8\).
   - The six-sided die has numbers: \(1, 2, 3, 4, 5, 6\).

2. **Determine the numbers that are multiples of 3:**
   - For the octahedral die: \(3, 6\).
   - For the six-sided die: \(3, 6\).

3. **Calculate the probability of rolling a multiple of 3 on each die:**
   - Probability of rolling a multiple of 3 on the octahedral die:
     \[
     P(\text{multiple of 3 on octahedral die}) = \frac{2}{8} = \frac{1}{4}
     \]
   - Probability of rolling a multiple of 3 on the six-sided die:
     \[
     P(\text{multiple of 3 on six-sided die}) = \frac{2}{6} = \frac{1}{3}
     \]

4. **Calculate the probability of rolling a number that is not a multiple of 3 on each die:**
   - Probability of not rolling a multiple of 3 on the octahedral die:
     \[
     P(\text{not a multiple of 3 on octahedral die}) = 1 - \frac{1}{4} = \frac{3}{4}
     \]
   - Probability of not rolling a multiple of 3 on the six-sided die:
     \[
     P(\text{not a multiple of 3 on six-sided die}) = 1 - \frac{1}{3} = \frac{2}{3}
     \]

5. **Calculate the probability that the product is not a multiple of 3:**
   - The product is not a multiple of 3 if neither roll is a multiple of 3:
     \[
     P(\text{product not a multiple of 3}) = P(\text{not a multiple of 3 on octahedral die}) \times P(\text{not a multiple of 3 on six-sided die})
     \]
     \[
     P(\text{product not a multiple of 3}) = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2}
     \]

6. **Calculate the probability that the product is a multiple of 3:**
   - The product is a multiple of 3 if at least one of the rolls is a multiple of 3:
     \[
     P(\text{product is a multiple of 3}) = 1 - P(\text{product not a multiple of 3}) = 1 - \frac{1}{2} = \frac{1}{2}
     \]

The final answer is \(\boxed{\frac{1}{2}}\).