Let $g(x)=\frac{1}{8} x^{3}+\frac{1}{2} x-\frac{1}{4}$ and let $h$ be the inverse function of $g$. Notice that $g(2)=\frac{7}{4}$.
\[
h^{\prime}\left(\frac{7}{4}\right)=
\]

Step 1 ：Let \(g(x)=\frac{1}{8} x^{3}+\frac{1}{2} x-\frac{1}{4}\) and let \(h\) be the inverse function of \(g\). Notice that \(g(2)=\frac{7}{4}\).

Step 2 ：We need to find \(h'(\frac{7}{4})\). The derivative of the inverse function can be found using the formula: \(h'(a) = \frac{1}{g'(h(a))}\) where \(g'(x)\) is the derivative of \(g(x)\) and \(h(a)\) is the inverse function of \(g(x)\) evaluated at \(a\).

Step 3 ：In this case, we need to find \(h'(7/4)\), so we need to find \(g'(h(7/4))\). But we know that \(g(2) = 7/4\), so \(h(7/4) = 2\). Therefore, we need to find \(g'(2)\).

Step 4 ：To find \(g'(2)\), we first need to find the derivative of \(g(x)\), \(g'(x)\).

Step 5 ：The derivative of \(g(x)\) at \(x=2\) is \(2\). Therefore, \(h'(7/4) = 1/g'(2) = 1/2\).

Step 6 ：Final Answer: \(\boxed{\frac{1}{2}}\).