Analytic in $\mathbb{C}$ implies $\left|\frac{f'(x)}{f(x)}\right|$ is bounded in $\mathbb{R}$? If $f(z)$ is an analytic function in the complex plane, $z=x+iy$, and $f(x)\neq 0$ for all $x\in \mathbb R$, does this imply that $\frac{f'(x)}{f(x)}$ is bounded on $\mathbb R$?i.e., $\big|\frac{f'(x)}{f(x)}\big|\leq C$, for some $C>0$.

The answer is negative.
For example $f(z)=\exp (z^2)$ is analytic and different from zero in the whole complex plane, and it has $f^\prime (z)=2z\ f(z)$. Hence for $x\in \mathbb{R}$ you get:
$$\left| \frac{f^\prime (x)}{f(x)}\right| =2|x|$$
which is not bounded from above on the real line.

I'd like to remark that "having a bounded logarithmic derivative" implies an exponential growth/decay estimate for $f(x)$.
In fact, assume you can find a function $f(z)$ which satifies your requirements, i.e. it is analytic in the whole plane, its restriction to the real line differs from zero everywhere and has bounded logarithmic derivative, i.e.:
$$\tag{BLD} \left| \frac{f^\prime (x)}{f(x)}\right| \leq C \qquad \text{, for }x\in \mathbb{R}$$
Assume for the time being also $f(x)>0$ for $x\in \mathbb{R}$ and $C>0$ (for, if $C=0$ then $f(x)$ is a constant); thus $f(x)$ satisfies the differential inequalities:
$$-C\ f(x)\leq f^\prime (x)\leq C\ f(x)$$
which imply the growth/decay estimates:
$$f(0)\ e^{-C|x|}\leq f(x)\leq f(0)\ e^{C|x|}\; .$$
If $f(x)<0$ then previous estimates rewrite:
$$f(0)\ e^{C|x|} \leq f(x)\leq f(0)\ e^{-C|x|}\; .$$
Therefore in any case your function $f(x)$ satisfies:
$$\tag{GDE} |f(0)|\ e^{-C|x|}\leq |f(x)|\leq |f(0)|\ e^{C|x|}\; .$$
Neverthless, I don't know if estimates (GDE) are equivalent to (BLD) in the case $f(x)$ is the restriction of an analytic function to the real line.
Certainly (GDE) is not equivalent to (BLD) for arbitrary real function: in fact, for example, the function $f(x) = \exp (|x|\ \sin x^4)$ is of class $C^1(\mathbb{R})$ (at least) and satisfies (GDE) with $C=1$, but it does not stisfy (BLD) for:
$$f^\prime (x) = \operatorname{sign}(x)\ f(x)\ (4\ x^4\ \cos x^4 + \sin x^4)$$
hence:
$$\left| \frac{f^\prime (x)}{f(x)}\right| = 4\ x^4\ \cos x^4 + \sin x^4 $$
which is not bounded.
