Obviously, it depends on what you mean by "piecewise differentiable." I'll show the following:
Claim: There exists a differentiable function $f: [0,1] \rightarrow \mathbb R$, such that if $R$ is the set where the function $\max\{f,0\}$ is differentiable, $R$ cannot be written as a countable union of intervals.
I'd argue that this $f$ is therefore not piecewise differentiable, for any reasonable definition of "piecewise differentiable".
Proof of Claim:
Step 1. Constructing $f$:
For $m \geq 0$, let $\mathcal A_m$ be the space of $m$-tuples $(a_1, ..., a_m)$ where each $a_i$ is either $0$ or $2$. If $A \in \mathcal A_m$, we define the open interval $I_A = (b_A + \frac 13 3^{-m}, b_A + \frac 23 3^{-m})$, where $b_A = \sum_{i=1}^m a_i 3^{-i}$. One can check that these intervals are disjoint.
The intervals $I_A$ for $A \in \mathcal A_m$ are precisely the open intervals removed in step $(m+1)$ in the standard construction of the Cantor set. So we can set
$$C = [0, 1] \setminus \bigcup_{m=0}^\infty \bigcup_{A \in \mathcal A_m} I_A\text;$$
this is then the standard Cantor set.
We will define $f$ as follows: If $x \in C$, then we set $f(x) = 0$. Otherwise, there is a unique $m \geq 0$ and a unique $A \in \mathcal A_m$ such that $x \in I_A$. We can write $I_A = (c_A, d_A)$, where $d_A - c_A = 3^{-m-1}$. Let $z(x) = 3^{m+1}(x - c_A)$, and let $f(x) = 3^{-2m-2} h(z(x))$, where $h(z) = z^2 (z-1)^2 (z-1/2)$. (The precise form of $h$ doesn't matter; the properties we need are that $h(0) = h'(0) = h(1) = h'(1) = h(1/2) = 0$, and $h'(1/2) \neq 0$.)
[A less formal but more readable definition of $f$: In each of the intervals that you remove to get a Cantor set, glue in a copy of $h$ above, where the values are rescaled according to the square of the length of the interval.]
Step 2. Check that $f$ is differentiable:
In the interior of each $I_A$, it is clear that $f$ is differentiable, since it is defined as a composition of differentiable functions. So we only need to check that $f$ is differentiable at each point in $C$. In fact, I claim that $f'(x) = 0$ whenever $y \in C$:
To show this, we will prove that $\lvert f(x)\rvert \leq (x-y)^2$ for all $x \in [0,1]$. To see this, suppose that $x \in I_A$, where $A \in \mathcal A_m$ for some $m$. (If not, then by definition $f(x) = 0$, and the inequality is obviously true.) Again, write $I_A = (c_A, d_A)$. Certainly, $\lvert x-y\rvert \geq d$, where $d = \min \{x-c_A, d_A - x\}$ is the distance from $x$ to the boundary of $I_A$. Referring to $z(x)$ from the definition of $f$, it follows that $\lvert x-y\rvert \geq 3^{-m-1} \min\{z(x), 1 - z(x)\}$. For $0 \leq z \leq 1$, it is easy to check that $\lvert h(z)\rvert \leq (\min \{z, 1-z\})^2$. Piecing everything together, the definition $f(x) = 3^{-2m-2} h(z(x))$ gives $\lvert f(x)\rvert \leq (x-y)^2$.
From this inequality, the definition of the derivative quickly shows that $f'(y) = 0$.
Step 3. The set where $\max\{f, 0\}$ is not differentiable:
Let $g = \max\{f, 0\}$. Let $S$ be the set of midpoints of the intervals $I_A$, for $A \in \bigcup_{m=0}^\infty \mathcal A_m$. I claim if $x \in S$, then $g$ is not differentiable at $x$.
This is pretty easy to check. In fact, if $x$ is the midpoint of $I_A$, where $A \in \mathcal A_m$, then using the chain rule, we can check that $f'(x) = 3^{-m-1} h'(1/2) = 3^{-m-1}/4$. This will be the right derivative of $g$ at $x$, whereas the left derivative will be zero, so $g$ is not differentiable at $x$.
Step 4. The set of points where $g$ is differentiable cannot be written as a countable union of intervals:
Suppose that $I$ is an interval where $g$ is differentiable. Then $I$ can contain at most one point of $C$, because between any two points of $C$, there must be one of the ""removed intervals" $I_A$, and therefore one point of $S$. Since the Cantor set $C$ is uncountable, this shows that $[0,1] \setminus S$ cannot be written as the union of such intervals.
