Let $f=\chi_\mathbb{Q}$ be the rational characteristic function (i.e., $f(x)=1$ if $x$ is rational, and $0$ otherwise). Is there any solution ($g,\eta$) for the equation $$ f(y)-f(x)=(y-x)g(\eta(x,y)), $$ where $g:\mathbb{R}\rightarrow \mathbb{R}$, and $\eta:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$ is a function with $\min\{x,y\}\leq \eta(x,y)\leq \max\{x,y\}$?
Note that $g\circ\eta(x,y)=0$, for all $(x,y)\in \mathbb{Q}\times \mathbb{Q}\cup \mathbb{Q}^c\times \mathbb{Q}^c$ with $x\neq y$, and so the set of zeros of $g$ is dense in $\mathbb{R}$.
Here's my idea of a solution (using Axiom of Choice and Zorn's lemma).
If $(x,y) \in Q\times Q \cup Q^c\times Q^c$ then let $\eta(x,y)$ be some rational number between $x$ and $y$, and let $g(t)=0$ if $t$ is rational. We might need Axiom of Choice to define this part of $\eta$.
Otherwise, i.e. when $(x,y) \in Q\times Q^c \cup Q^c\times Q$, I think that we can refer to cardinality and then use Zorn's lemma to conclude that there is a bijection $\eta : Q\times Q^c \cup Q^c\times Q \to Q^c$ such that $\min(x,y) \leq \eta(x,y) \leq \max(x,y)$. Then let $g(t) = d(\eta^{-1}(t))$, where $d(x,y) = (f(y)-f(x))/(y-x)$, if $t$ is irrational.
EDIT 14 May 2017
Let $C \subset [0, 1]$ be the ordinary Cantor set. Then let $S = C + Q = \{ c+q \, | \, c \in C, \, q \in Q \}$. This set is very dense; for every open interval $U \subseteq R$, the intersection $S \cap U$ contains an uncountable infinity of points. Still, $m(S) = 0$, where $m$ is the Lebesgue measure. Therefore there must be an uncountable infinity of sets $S_\alpha$ that (in some sense) are similar to $S$ and such that $S_\alpha \cap S_\beta = \emptyset$ whenever $\alpha \neq \beta$.
Since $Q \times Q^c \cup Q^c \times Q$ and $\{ S_\alpha \}$ have the same cardinality, there's a bijection $\phi : Q \times Q^c \cup Q^c \times Q \to \{ S_\alpha \}$. Using the Axiom of Choice we can now find $\eta : Q \times Q^c \cup Q^c \times Q \to Q^c$ such that $\eta(x,y) \in \phi(x,y) \cap [\min(x,y), \max(x,y)]$. By construction $\eta$ is injective.
I did some handwaving about the existence of the $S_\alpha$ and will therefore continue to think about how those can be constructed from $S$.