It is recommended and nearly required to write some of your thoughts and attempts. And perhaps provide more context.

@JacobsonRadical Done

Have you tried just checking using the definition of a distributional derivative?

@Deane I've edited the question recalling the definition of distributional derivative. But I don't see how to get an answer to the question(s) from it.

Stick in the specific function $f$ into the formula and see what you get.

@Deane Done (see new edit), but the question(s) remain

You could start with some simpler examples. For example, suppose $$u(x) = \begin{cases} 1 &\text{ if }0 \le x \le 1\\ 0 &\text{ otherwise} \end{cases}. $$ What is $$\int u(x)f(x)\,dx?$$ Next, let $u$ be any compactly supported smooth function. What is $$\int u(x)f(x)\,dx?$$

@Deane I'm very sorry, but I don't see where this leads.

@Deane $$\int_{\mathbb R} u(x) f(x) dx =-\int_{\mathrm{supp} u \cap (\mathbb R\setminus \mathbb Q)} u(x) dx$$ right?

How about the first integral?

@Deane $$=-\int_{[0,1]} ({1}_{\mathbb R \setminus \mathbb Q}) = - 1, $$ right?

Yes! Now, could you revise your answer to the second integral?

I mean, simplify it.

Do you mean that $$ -\int_{\mathrm{supp} u \cap (\mathbb R \setminus \mathbb Q)} u(x) dx = -\int_{\mathrm{supp} u} u(x) dx $$?

Yup. Now go back to the definition of the distributional derivative of the function $f$. See if you can figure out what the answer is. If not, just write out the formula and ask again.

By the way, the fact you're using (but maybe initially forgot) is that if $f = g$ a.e., then $\int f = \int g$.

@Deane This is weird: $$\langle T_f ', \varphi \rangle = - \int_{\mathbb R} \varphi'(x) dx $$

Good!. Now remember that $\phi$ is compactly supported and therefore so is $\phi'$. So it suffices to integrate over a compact interval that contains the support of $\phi'$. Then use the fundamental theorem of calculus.

@Deane Are we saying that $T'_f = 0$?

Well, you seemed to have provided a pretty ironclad proof of that.

@Deane This is very surprising though. Despite the fact that $f$ is nowhere continuous, its devative behaves like the derivative of a constant?

Here's are some basic facts you can prove yourself: If a function $g$ is differentiable, then its distributional derivative is equal to its derivative. If $f = g$ a.e., then the distributional derivatives of $f$ and $g$ are equal.

Two things: 1) I agree that the distributional (weak) derivative is at first a strange thing. 2) It's great to think about examples like the one you posted. But always try to compute the answer yourself using just the definitions used and, if necessary, some basic facts you already know. It's often easier than you might think at first.