Let $h(x)=5-x-x^{3}$ and let $f$ be the inverse function of $h$. Notice that $h(-1)=7$. \[ f^{\prime}(7)=\square \]

Step 1 ：Let \(h(x)=5-x-x^{3}\) and let \(f\) be the inverse function of \(h\). Notice that \(h(-1)=7\).

Step 2 ：We need to find \(f'(7)\). To do this, we can use the formula for the derivative of an inverse function, which is \(1/(f'(f^{-1}(x)))\).

Step 3 ：Here, we know that \(h(-1) = 7\), so \(f(7) = -1\).

Step 4 ：We need to find \(h'(x)\) and then substitute \(x = -1\) into it to find \(f'(7)\).

Step 5 ：First, find the derivative of \(h(x)\), which is \(h'(x) = -3x^{2} - 1\).

Step 6 ：Then, substitute \(x = -1\) into \(h'(x)\) to get \(h'(-1) = -4\).

Step 7 ：Finally, substitute \(h'(-1)\) into the formula for the derivative of an inverse function to get \(f'(7) = -1/4\).

Step 8 ：Final Answer: \(\boxed{-\frac{1}{4}}\)