Instead of looking at the case where the die has been rolled 9 times, consider the case where the die has only been rolled once.
Let $X_{1,i}$ be an "indicator" for the die rolling a 1 on the $i$th roll. Put explicitly:
$$
X_{1,i} =
\begin{cases}
1, \ \ \text{die rolls 1 on $i$th roll}\\
0, \ \ \text{die does not roll 1 on $i$th roll}\\
\end{cases}
$$
Similarly, let $X_{6,i}$ be the indicator variable for the die rolling a 6 on the $i$th roll:
$$
X_{6,i} =
\begin{cases}
1, \ \ \text{die rolls 6 on $i$th roll}\\
0, \ \ \text{die does not roll 6 on $i$th roll}\\
\end{cases}
$$
What is the covariance between $X_{i,1}$ and $X_{6,i}$? We can use the decomposition formula:
$$
\text{Cov} (X_{1,i},X_{6,i}) = E(X_{1,i}*X_{6,i}) - E(X_{1,i})E(X_{6,i})
$$
$E(X_{1,i})$ and $E(X_{6,i})$ are easy since $X_{1,i}$ are Bernoulli random variables with probability 1/6. Their expected variables will both be 1/6.
To calculate $E(X_{1,i}*X_{6,i})$, note that the random variable $Z = X_{1,i} *X_{6,i}$ equals 0 with probability 1.
If the $i$th roll is 1, $Z = 1*0 = 0$, if the $i$th roll is 6, $Z = 0*1 = 0$, and if the $i$th roll is neither 1 nor 6, then $Z = 0*0 = 0$. Therefore, the expected value of Z is 0.
$$
E(Z) = E(X_{1,i} X_{6,i}) = 0
$$
Putting into the formula above,
$$
\text{Cov} (X_{1,i},X_{6,i}) = E(X_{1,i}*X_{6,i}) - E(X_{1,i})E(X_{6,i})
$$
$$
= 0 - (1/6)(1/6) = -1/36
$$
Now, that is for just one roll, and I hope it's a good start to you solving this. To finish this problem, note:
(1) What is be the relationship between $X_{i,1}$, $X_{i,6}$, $S(1)$, and $S(6)$?
(2) Look into the summation formulas for Covariance
(3) The throws are independent, so Cov$(X_{1,i},X_{6,j}) = 0$ for $i \neq j$.
Once you have the covariance, the correlation can be calculated in a straightforward by noting that $S(1)$ and $S(6)$ are both Binomial random variables (information that can be used to get the variances, which are then used to scale the covariance).
