Use the change-of-base theorem to find the logarithm.
\[
\log _{1 / 6} 3
\]

Step 1 ：We are given the logarithm \(\log _{1 / 6} 3\) and we are asked to find its value.

Step 2 ：We can use the change-of-base theorem to solve this problem. The change-of-base theorem states that for any positive numbers a, b, and c (where a ≠ 1 and b ≠ 1), the logarithm base b of a can be computed with the formula: \(\log_b a = \frac{\log_c a}{\log_c b}\)

Step 3 ：In this case, we want to find the value of \(\log _{1 / 6} 3\), so we can use the change-of-base theorem with a = 3, b = 1/6, and c = 10 (since the base 10 logarithm is commonly used and easy to compute).

Step 4 ：So, we need to compute \(\frac{\log_{10} 3}{\log_{10} (1/6)}\).

Step 5 ：Let's calculate the values of \(\log_{10} 3\) and \(\log_{10} (1/6)\).

Step 6 ：\(\log_{10} 3\) is approximately 0.47712125471966244 and \(\log_{10} (1/6)\) is approximately -0.7781512503836436.

Step 7 ：Substitute these values into the formula, we get \(\frac{0.47712125471966244}{-0.7781512503836436}\) which is approximately -0.6131471927654584.

Step 8 ：Final Answer: The value of \(\log _{1 / 6} 3\) is \(\boxed{-0.6131471927654584}\).