$\begin{array}{l}f x=\sqrt[3]{1 x-3} \\ f^{-1}(x)=\end{array}$

Step 1 ：Given the function \(f(x) = \sqrt[3]{1x - 3}\), we want to find the inverse function \(f^{-1}(x)\).

Step 2 ：To find the inverse function, we first replace \(f(x)\) with \(y\), so we have \(y = \sqrt[3]{1x - 3}\).

Step 3 ：Next, we swap \(x\) and \(y\) to get \(x = \sqrt[3]{1y - 3}\).

Step 4 ：Now, we solve this equation for \(y\). First, we cube both sides to get rid of the cube root: \(x^3 = 1y - 3\).

Step 5 ：Then, we add 3 to both sides to isolate \(y\): \(x^3 + 3 = y\).

Step 6 ：So, the inverse function is \(f^{-1}(x) = x^3 + 3\).

Step 7 ：Finally, we check our result by substituting \(f^{-1}(x)\) into the original function \(f(x)\) and simplifying. If we get \(x\), then our result is correct.

Step 8 ：Substituting \(f^{-1}(x)\) into \(f(x)\), we get \(f(f^{-1}(x)) = \sqrt[3]{1(f^{-1}(x)) - 3} = \sqrt[3]{1(x^3 + 3) - 3} = \sqrt[3]{x^3 + 3 - 3} = \sqrt[3]{x^3} = x\).

Step 9 ：Since \(f(f^{-1}(x)) = x\), our result is correct. So, the inverse function is \(\boxed{f^{-1}(x) = x^3 + 3}\).