Find the derivative of the function.
\[
\begin{array}{r}
s(x)=\frac{e^{6 x-1}}{x^{3}-1} \\
s^{\prime}(x)=\square
\end{array}
\]

Step 1 ：Let \(f(x) = e^{6x-1}\) and \(g(x) = x^3 - 1\).

Step 2 ：Find the derivative of \(f(x)\) and \(g(x)\), denoted as \(f'(x)\) and \(g'(x)\) respectively.

Step 3 ：\(f'(x) = 6e^{6x-1}\) and \(g'(x) = 3x^2\).

Step 4 ：Substitute \(f(x)\), \(g(x)\), \(f'(x)\), and \(g'(x)\) into the quotient rule formula: \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}\).

Step 5 ：Thus, the derivative of the function \(s(x)=\frac{e^{6 x-1}}{x^{3}-1}\) is \(s^{\prime}(x)=\frac{-3x^{2}e^{6x-1} + 6(x^{3} - 1)e^{6x-1}}{(x^{3} - 1)^{2}}\).

Step 6 ：\(\boxed{s^{\prime}(x)=\frac{-3x^{2}e^{6x-1} + 6(x^{3} - 1)e^{6x-1}}{(x^{3} - 1)^{2}}}\)