Yes, it implies that $f$ is differentiable at $x=0$, with f'(0) = \mathop {\lim }\limits_{x \to 0} f'(x). This can be easily proved using the mean-value theorem.
EDIT:
Proof: Let l=\lim _{x \to 0} f'(x). Let us first show that $ \mathop {\lim }\limits_{h \to 0^ + } \frac{{f(h) - f(0)}}{h} = l. $ Let $\varepsilon > 0$. Then, since \lim _{x \to 0^+} f'(x) = l, there exists $\delta > 0$ such that |f'(x) - l| < \varepsilon for any $x \in (0,\delta)$. Suppose next that $h \in (0,\delta)$. Since $f$ is continuous on $[0,h]$ and differentiable on $(0,h)$, by the mean-value theorem there exists $c \in (0,h)$ such that f'(c) = \frac{{f(h) - f(0)}}{h}. Noting that $c \in (0,\delta)$, we thus have |f'(c) - l| < \varepsilon. So, given any $\varepsilon > 0$, there exists $\delta > 0$ such that $h \in (0,\delta)$ implies $ \bigg|\frac{{f(h) - f(0)}}{h} - l \bigg| < \varepsilon. $ Thus, by definition of (one-side) limit, we have $ \mathop {\lim }\limits_{h \to 0^ + } \frac{{f(h) - f(0)}}{h} = l. $ Analogously, $ \mathop {\lim }\limits_{h \to 0^ - } \frac{{f(h) - f(0)}}{h} = l. $ Hence $ \mathop {\lim }\limits_{h \to 0 } \frac{{f(h) - f(0)}}{h} = l, $ that is f'(0) = l.