$f$ analytic on $B(R,z_0)$ with $z_0$ on the real axis and $\forall\zeta\in\mathbb{R}\cap B(R,z_0),\Im f(\zeta)=0$ then $\partial _z f (\zeta) = 0$ If $f$ is analytic on $B(R,z_0)$ with $z_0 $ on the real axis and if $ \forall \zeta \in \mathbb{R} \cap B(R,z_0), \Im f(\zeta) = 0  $ then $ \partial _z f (\zeta) = 0 $.
I can write $f = u + iv$ with $u,v : \mathbb{C} \rightarrow \mathbb{R} \leadsto \partial _z f = \frac{1}{2} (\partial_x - i \partial_y) (u + iv) \stackrel{\text{Cauchy-Riemann}}{=} (\partial_y + i \partial_x) v $.
But since $\Im f(\zeta) = 0 $ then $ v(\zeta) = 0 $ and so $\partial _z f (\zeta) = 0$.
Is this true? I don't think so but can't figure why...
 A: $f(z)=z$ is a counterexample. 
The mistake is in concluding from $v(\zeta)=0$ that $\partial _z v(\zeta)=0$. A function of two variables may well vanish on a line but have a nonzero gradient there: $v(x,y)=y$ is an example of such function.
A: No, this is false: the function  $f(z)=z$ gives a counterexample, since for all $\zeta \in \mathbb R$ you have $Im(f)(\zeta)=0$ but nevertheless $(\partial_zf)(\zeta)=1$ for those $\zeta$.  
Your argumentation in the general case is correct up to the equality $\partial _z f = (\partial_y + i \partial_x) v $.
But you may not argue that since $v(\zeta)=0$ then $[(\partial_y + i \partial_x) v](\zeta)=0$.
This is like falsely arguing that 
"Since $\frac {\operatorname d}{\operatorname dx}sin =cos$ and since $\operatorname {sin}(0)=0$, then  $ \frac {\operatorname d}{\operatorname dx}sin(0)=cos(0)=0$"  
The gist of the matter is that in any context you care to consider, there is no relation between the value of a function at a point and that of some derivative of  that function at that same  point.
