The problem is as follows:
Let $f: \mathbb{Q} \to \mathbb{Q}$ and consider the differential equation $f' = f$, with the standard definition of differentiation. Do there exist any nontrivial solutions?
(Note that of course $f \equiv 0$ is a solution - by "nontrivial solutions", I mean anything else).
Observations:
Differentiation and continuity are much weaker concepts on the rationals. For example, $H(x-\sqrt{2}) : \mathbb{Q} \to \mathbb{Q}$ is continuous and everywhere differentiable, where $H$ is the Heaviside step function.
If there exists a nontrivial solution $f_0$, then there are uncountably many solutions. For example, $H(x-\alpha)f_0$ is also a solution for any irrational $\alpha$ (which already gives uncountably many solutions), and thus any* linear combination $k_0 f_0 + \sum_{\alpha \in A} H(x-\alpha)k_\alpha f_0$ (with $A \subset \mathbb{R}\setminus\mathbb{Q}$) is also a solution, by linearity of the DE.
We can answer in the negative if there is a way to show that any such solution could be extended to a solution to $f' = f$ on $\mathbb{R}$, because those solutions are simply $ke^x$, which takes irrational values over the rationals unless $k = 0$. Unfortunately, the solutions $f : \mathbb{Q} \to \mathbb{Q}$ for the a differential equation $\mathcal{L}y = 0$ are not, in general, a subset of the real solutions. e.g. $y' = 0$ has solution $H(x-\sqrt{2})$ but every solution on $\mathbb{R}$ must be constant.
If the answer is "yes", maybe we'd hope to be able to construct a solution via some iterative method, but since Cauchy sequences are not in general convergent, we'd need some sort of machinery to guarantee rational limits.
*you can either insert the word "finite" here, or stipulate that the $k_\alpha$ are such that the quantity $\sum_{\alpha<q} k_\alpha$ is finite and rational for all $q \in \mathbb{Q}$, but the point is that we can construct a bunch of "different looking" solutions.