The general formula for the \( n \)-th roots of a complex number \( z = r \text{cis} \theta \) is given by:
\[
z_k = \text{cis} \left( \frac{\theta + 2k\pi}{n} \right), \quad k = 0, 1, 2, \dots, n-1
\]
For \( -1 \), we have \( r = 1 \) and \( \theta = \pi \). We need to find one of the 15th roots of \( -1 \), so we apply the formula with \( n = 15 \):
\[
z_k = \text{cis} \left( \frac{\pi + 2k\pi}{15} \right)
\]
For \( k = 6 \), we get:
\[
z_6 = \text{cis} \left( \frac{\pi + 2(6)\pi}{15} \right) = \text{cis} \left( \frac{13\pi}{15} \right)
\]
Thus, the correct answer is \( \text{cis} \frac{13\pi}{15} \), which corresponds to option (3).