If it's differentiable at every point, then this can't happen. This follows from the mean value theorem:
If $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then for at least one point $c$ between $a$ and $b$, we have f'(c) = \dfrac{f(b)-f(a)}{b-a}.
If your $f(x)$ is not constant, but is differentiable everywhere, pick an $a$ and $b$ with $f(a)\neq f(b)$. By the MVT, we have f'(c) = \dfrac{f(b)-f(a)}{b-a} \neq 0 since $f(b) \neq f(a)$.
On the other hand, if you assert your function is differentiable only "almost everywhere" instead of "everywhere" (in a sense which can be made precise) and that the derivative "almost everywhere" is equal to $0$, then this can happen. The standard example is Cantor's function (also known as the Devil's Staircase). See http://en.wikipedia.org/wiki/Cantor_function.