Whether there exists a continuos monotone function $f: [0,1]\rightarrow [0,1]$ with the following properties:
(1) $f$ strictly increase, $f(0)=0$ and $f(1)=1$;
(2) there is no interval $A\subset [0,1]$, where the derivative $f'$ (a) exists, (b) is continuous and (c) is finite;
(3) there exists a set $B\subset [0,1]$ dense in $[0,1]$, where the derivative $f'$ (a) exists, (b) is positive and (c) is finite?
I can construct an example of function $f:[0,1]\rightarrow [0,1]$ with condition (1), whose derivative is either $0$, or $+\infty$ (at points, where the dewrivative exists). It follows from Lebesgues theorem that $f'$ is zero on some set $A\subset[0,1]$, which has Lebesgue measure 1 and, whence, is dence in $[0,1]$. Clearly, since the derivative is either $0$, or $+\infty$ (in the example, which I can construct), and the function is invertible, then $f'\neq 0$ at any interval.
In other words, I can answer "yes" to my question if ommit the word "positive" in the condition (3).
I hope that my question is "natural", but neither have found "in the internet" neither its proof, nor counter example, nor this question as it is (for example as open problem).
The answer is yes, see Pompeiu derivative. Short explanation : denote $(q_j)_{j \in \mathbb{N}}$ an enumeration of all rationals in $[0,1]$, and denote $$g : x \mapsto \sum \limits_{j \in \mathbb{N}} 2^{-j} \sqrt[3]{x-q_j}, \ \mbox{ and }\ f : x \mapsto \frac{g(x)-g(0)}{g(1)-g(0)}$$
It is easy to see that $g$ is well defined and strictly increasing, so $f$ is well-defined, strictly increasing and $f(0)=0$, $f(1)=1$. It is clear that if we want $f$ satisfying $(2)$ and $(3)$, it is enough to check that $g$ statisfies $(2)$ and $(3)$.
$(2)$ is obvious : for every $j \in \mathbb{N}$, $g'(q_j) = +\infty$, and $\{q_j, j\in \mathbb{N}\}$ is dense in $[0,1]$.
Let us prove $(3)$. We consider some open interval $I = ]x_0-\varepsilon,x_0+\varepsilon[ \subset [0,1]$, and prove that that $B \cap I \neq \varnothing$. It is easy to show that $g'(x) = \frac{1}{3}\sum \limits_{j \in \mathbb{N}} \frac{2^{-j}}{\sqrt[3]{(x-q_j)^2}}$ is finite at every point where this sum is finite. The measure of $\tilde{B} := [0,1] \backslash \bigcup \limits_{j \in \mathbb{N}} \big[ q_j-\frac{\varepsilon}{2^{j+2}}, q_j+\frac{\varepsilon}{2^{j+2}}\big]$ is $|\tilde{B}| \ge 1 - \sum \limits_{j=0}^{\infty} \frac{2\varepsilon}{2^{j+2}} = 1- \varepsilon$. Thus $\tilde{B}$ and $I$ cannot be disjoint. For $x \in \tilde{B} \cap I$, $\sum \limits_{j \in \mathbb{N}} \frac{2^{-j}}{\sqrt[3]{(x-q_j)^2}} \le \sum \limits_{j \in \mathbb{N}} \frac{1}{\sqrt[3]{\varepsilon}} \frac{2}{\sqrt[3]{2}}\big)^{-j} < +\infty$ because $\frac{2}{\sqrt[3]{2^2}} < 1$, so $B \cap I \neq \varnothing$
I think there is even a function $f$ continuous, satisfying $(1)$, $(2)$, $(3)$ and such that $f$ has a finite derivative everywhere (but unbounded on every interval because of 2). It would require more effort to build than what is above.