No, such functions do not exist. More precisely, let $f:\mathbb R\to\mathbb R$ be an arbitrary function, $\Sigma$ is the set of $x\in\mathbb R$ such that $f'(x)$ exists and equals 0. Then $f(\Sigma)$ has measure 0.

By countable subadditivity of measure, we may assume that the domain of $f$ is $[0,1]$ rather that $\mathbb R$. Fix an $\varepsilon>0$. For every $x\in\Sigma$ there exists a subinterval $I_x\ni x$ of $[0,1]$ such that $f(5I_x)$ is contained in an interval $J_x$ with $m(J_x)<\varepsilon m(I_x)$. Here $m$ denotes the Lebesgue measure and $5I_x$ the interval 5 times longer than $I_x$ with the same midpoint. Now by Vitali Covering Lemma there exists a countable collection $\{x_i\}$ such that the intervals $I_{x_i}$ are disjoint and the intervals $5I_{x_i}$ cover $\Sigma$. Since $I_{x_i}$ are disjoint, we have $\sum m(I_{x_i})\le 1$. Therefore $f(\Sigma)$ is covered by intervals $J_{x_i}$ whose total measure is no greater than $\varepsilon$. Since $\varepsilon$ is arbitrary, it follows that $f(\Sigma)$ has measure 0.