$\lim _{h_{-} \rightarrow 0} \frac{(1+h)^{\frac{1}{6}}-1}{h}$

Step 1 ：First, we recognize that the limit is in the form of \(\frac{0}{0}\), which is an indeterminate form. This suggests that we can use L'Hopital's Rule.

Step 2 ：According to L'Hopital's Rule, when the limit is in the form of \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can differentiate the numerator and the denominator separately and then take the limit.

Step 3 ：So, we differentiate the numerator and the denominator separately. The derivative of \((1+h)^{\frac{1}{6}}\) with respect to \(h\) is \(\frac{1}{6}(1+h)^{-\frac{5}{6}}\), and the derivative of \(h\) with respect to \(h\) is 1.

Step 4 ：Substitute these derivatives back into the limit, we get \(\lim _{h_{-} \rightarrow 0} \frac{\frac{1}{6}(1+h)^{-\frac{5}{6}}}{1}\).

Step 5 ：As \(h\) approaches 0, \((1+h)^{-\frac{5}{6}}\) approaches 1. So, the limit is \(\frac{1}{6}\).

Step 6 ：Thus, the final answer is \(\boxed{\frac{1}{6}}\).