The differentiable-functions extension of the binomial product identity

Let $n$ be a natural number, and let $f$ and $g$ be differentiable functions. Differentiable-functions identity.

$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{f(n)k+g(n)}=n^n n!\prod_{k=0}^{n}\frac{1}{f(n)k+g(n)}.$$

The preceding identities establish special cases involving linear expressions in $n$ and $k$, while the paper suggests investigating this broader formulation; no proof or resolution is supplied here.