Let $f$ be a continuous function. The quadratic variation of $f$ on $[a,b]$ is the usual definition: if $S$ is the set of Riemann-style partitions of $[a,b]$ directed by refinement, then the quadratic variation is the net-theoretic limit $$\int_a^b{\mathcal{Q}[df(t)]}=\lim_S{\sum_{[x,y]\in S}{(f(y)-f(x))^2}}$$ Varying $[a,b]$, we get a measure on Borel subsets of $\mathbb{R}$ by Carathéodory extension; I will call it $\mathcal{Q}_f$.
If $f$ is differentiable on $[a,b]$, then $\mathcal{Q}_f([a,b])=0$. In fact, since the closed subintervals of $[a,b]$ are dense in the measure algebra of $[a,b]$, $\mathcal{Q}_f$ is unsupported on $[a,b]$. But what if $f$ is only differentiable on some $S\subseteq[a,b]$?
I would like to say something like:
If $f$ is differentiable on $S$, then $S\cap\text{supp}{\mathcal{Q}_f}=\emptyset$.
Is this true? I will settle for the weaker claim that $S$ and $(\text{supp}{\mathcal{Q}_f})^{\mathsf{c}}$ have the same Lebesgue measure. You are also free to assume that $f$ is sufficiently regular as to make $\mathcal{Q}_f$ absolutely continuous (w.r.t. Lebesgue measure), although I would hope that such an assumption is not necessary.
"Brute force" does not work
If $f$ is differentiable on a nowhere dense set (e.g. $f$ is diffentiable on Fat Cantor dust), then the most we can say is that, if $p\in S$, then $|f(y)-(f(p)+f'(p)(y-p))|<\epsilon|y-p|$ near $p$. But then, if we try to estimate a partition on $[p-\delta,p+\delta]$, we can have arbitrarily-many points near $p+\delta$, each of which contributes $\approx\epsilon^2\delta^2$ to the sum.
Abstract measure theory does not work
Since Lebesgue measure is regular, then there is some compact set $K\subseteq S$ of measure $|S|-\epsilon$. If we could write $K$ as a union of countably-many compact intervals, then we should be done. But of course only open sets can be written as a union of countably-many open intervals.
Since we may assume $\mathcal{Q}_f$ absolutely continuous, $\mathcal{Q}_f$ has some density $q_f$. Let $x$ be a point of differentiability. Then for all $x$ except possibly a set of Lebesgue measure $0$, \begin{align*} q_f(x)&=\lim_{b,a\to x}{\frac{1}{b-a}\int_a^b{\mathcal{Q}[df]}} \\ &=\lim_{b,a\to x}{\lim_{S\to[a,b]}{\sum_{[x,y]\in S}{\frac{(f(y)-f(x))^2}{b-a}}}} \\ &=\lim_{b,a\to x}{\lim_{N\to\infty}{\frac{1}{N}\sum_{k=0}^{N-1}{\frac{\left(f\left(\left(1-\frac{k+1}{N}\right)a+\left(\frac{k+1}{N}\right)b\right)-f\left(\left(1-\frac{k}{N}\right)a+\left(\frac{k}{N}\right)b\right)\right)^2}{\frac{b-a}{N}}}}} \\ &\geq\lim_{b,a\to x}{\lim_{N\to\infty}{\inf_{0\leq\beta-\alpha\leq\frac{b-a}{N}}{\frac{(f(\beta)-f(\alpha))^2}{\beta-\alpha}}}} \\ &=\lim_{b,a\to x}{\inf_{0\leq\beta-\alpha\leq b-a}{\frac{(f(\beta)-f(\alpha))^2}{\beta-\alpha}}} \\ &=\liminf_{b,a\to x}{\frac{(f(b)-f(a))^2}{b-a}} \\ &=\lim_{b,a\to x}{\frac{f(b)-f(a)}{b-a}\cdot(f(b)-f(a))} \\ &=f'(x)\cdot0 \end{align*} Likewise for the sup.