Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ a differentiable function such that $f'(x)=0$ for all $x\in\mathbb{Q}$ Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ a differentiable function such that $f'(x)=0$ for all $x\in\mathbb{Q}.$ $f$ is a constant function?
 A: No, such a function is not necessarily constant.
At the bottom of page 351 of Everywhere Differentiable, Nowhere Monotone Functions, Katznelson and Stromberg give the following theorem:

Let $A$ and $B$ be disjoint countable subsets of $\mathbb{R}$.  Then there exists an everywhere differentiable function $F: \mathbb{R} \to \mathbb{R}$ satisfying
  
  
*
  
*$F'(a) = 1$ for all $a \in A$,
  
*$F'(b) < 1$ for all $b \in B$,
  
*$0 < F'(x) \leq 1$ for all $x \in \mathbb{R}$.
  

Choosing $A = \mathbb{Q}$ and an arbitrary (nonempty*) countable set $B \subset \mathbb{R} \setminus \mathbb{Q}$, we get an everywhere differentiable function $F$ with $F'(q) = 1$ when $q \in \mathbb{Q}$.  So if we define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = F(x) - x$, this then is the desired function satisfying $f'(q) = F'(q) - 1 = 1 - 1 = 0$ for $q \in \mathbb{Q}$, and $f$ is not constant (or else its derivative would be zero everywhere by the mean value theorem).
*(in case you take "countable" to mean either "countably infinite" or "finite")
