$\def\FF{\mathbb{F}}$I'm just guessing, but I would have thought it was the following: Hilbert reciprocity for function fields can be deduced from Weil reciprocity. Weil reciprocity is the following statement: Let $X$ be a complete curve over an algebraically closed field $k$. For any point $x \in X$ and nonzero meromorphic functions $f$ and $g$, define $(f,g)_x = (-1)^{(\mathrm{ord}_x f)(\mathrm{ord}_x g)}(f^{\mathrm{ord}_x g}/g^{\mathrm{ord}_x f})(x)$. Then $\prod_{x \in X} (f,g)_x=1$.
See here and here for the connection.
Now, over $\mathbb{C}$, we can prove Weil reciprocity as follows: Choose a path $\delta$ connecting $0$ to $\infty$ in $\mathbb{CP}^1$ and avoiding the critical values of $f$. For simplicity, let us assume $f$ has simple zeroes and poles $\zeta^{\pm}_1$, $\zeta^{\pm}_2$, ..., $\zeta^{\pm}_n$. Set $\gamma = f^{-1}(\delta)$. Then $\gamma$ is the union of $\deg(f)$ closed line segments. After reordering, we may assume $\zeta^+_i$ is joined to $\zeta^-_i$, say by $\gamma_i$.
We can define $\log(f)$ on $X \setminus \gamma$, by composing $f$ with a branch of $\log$ on $\mathbb{CP}^1 \setminus \delta$. The differential form $\omega:= \tfrac{1}{2 \pi i} \log(f) \tfrac{dg}{g}$ therefore makes sense on $X \setminus (\gamma \cup g^{-1}(\{ 0,\infty \}))$. If we integrate $\omega$ on little contours around the zeroes and poles of $g$, we get $\sum_{x \in X} \mathrm{ord}_x(g) \log(f(x))$.
On the other hand, if we integrate around a tubular neighborhood of $\gamma_i$, we pick up $\int_{\gamma_i} \tfrac{dg}{g} = \log(g(\zeta^{+}_i) - \log(g(\zeta^-_i))$ for some branch of $\log$. Summing on $i$, this is
$\sum_{x \in X} \mathrm{ord}_x(f) \log(g(x))$
The sum of the contours around the zeroes of $f$ is homologous to the sum over the neighborhoods of the $\gamma_i$, so we deduce
$$\sum_{x \in X} \mathrm{ord}_x(g) \log(f(x)) = \sum_{x \in X} \mathrm{ord}_x(f) \log(g(x))$$
and exponentiating gives the result.
A resource might be
Maurice Lecat, Erreurs de mathématiciens des origines à nos jours, Bruxelles : Castaigne, 1935
Cheers, Scott
Best Answer
In Index to Mathematical Problems, 1980-1984 - Page xi by Stanley Rabinowitz this quote is attributed to Léo Sauvé: