My turn:
First the BV part. Let t_0=0 < t_1<... be a partition of $[0,1]$. As mentioned in another proof, we can always add irrational points $t_i'$ in between to refine the partition. Then we have t_0 . Since $f(t_i') = 0$, we can form the estimate:
$\sum_{i=0}^{n-1} | f(t_{i+1}) - f(t_i)| \leq \sum_{i=0}^{n-1} | f(t_{i+1}) - f(t_i)'|+|f(t_i')-f(t_i)| \leq 2 \sum_{i=0}^{n} f(t_i)$
The last estimate can be replaced by summing over all points in $[0,1]$ where $f$ is positive: $\sum_{i=0}^{n-1} | f(t_{i+1}) - f(t_i)| \leq \sum_{t \in [0,1], f(t)>0} f(t) \leq \sum_{n=1}^{\infty} \frac{2(n+1)}{n^3} = K$ (The final inequality comes from the fact that there are at most $n+1$ rationals with denominator $n$ in $[0,1]$.) Consequently, we have $\mathrm{Var}_{[0,1]} f \leq K$.
Now the differentiability part. It turns out that $f$ is differentiable at all irrational points. To see this, let $\alpha \in (0,1)$ be irrational, and choose $y \in [0,\alpha)$. (The analysis for $y \in (\alpha, 1]$ is similar, mutatis mutandis.)
I will prove differentiability in two parts, the first part is unnecessary, but motivates the slightly more involved second part.
The key aspect here is that there are at most $n |y-\alpha|$ (not $n|y-\alpha|+1$!!!) rationals with denominator $n$ in $[y,\alpha]$. This is because $\alpha$ is irrational. Following a similar line of reasoning to above, we can replace the $n+1$ by $n |y-\alpha|$ to get the estimate:
$\mathrm{Var}_{[y,\alpha]} f \leq \sum_{n=1}^{\infty} \frac{2n|y-\alpha|}{n^3} \leq K|y-\alpha|$ We are almost finished, the only issue is to replace $K$ by an arbitrary $\epsilon>0$. Pick an $\epsilon > 0$, and choose $N$ such that \sum_{n=N+1}^{\infty} \frac{2n}{n^3} < \epsilon. Now choose $\delta>0$ such that the interval $(\alpha-\delta, \alpha+\delta)$ contains no rational of the form $\frac{k}{n}$, where $k \leq n \leq N$. (Such a $\delta$ must exist since $\alpha$ is irrational.) Then if $y \in (\alpha-\delta, \alpha+\delta)$, we can repeat the above analysis to get the estimate:
$\mathrm{Var}_{[y,\alpha]} f \leq \sum_{n=N+1}^{\infty} \frac{2n}{n^3} |y-\alpha| \leq \epsilon|y-\alpha|$ Finally, since $|f(y)-f(\alpha)| \leq \mathrm{Var}_{[y,\alpha]} f$, we have that $f'(\alpha)$ exists and is $0$.