Does $f'(x)$ always remain close to $f(x)/x$ as $x \rightarrow 0$?

Here is a counterexample: Define

$$\tag 1 f(x) = \sum_{k=1}^{\infty}\frac{x^{1/k}}{2^k},$$

the series converging uniformly on $[0,1].$ Each $x^{1/k}$ can be written

$$x^{1/k} = 1-\sum_{n=1}^{\infty}c_k(n)(1-x)^n,$$

where $0\le c_k(n)\le 1$ for all $k,n.$ Thus

$$f(x) = 1 - \sum_{k=1}^{\infty}\frac{1}{2^k}\sum_{n=1}^{\infty}c_k(n)(1-x)^n = 1 - \sum_{n=1}^{\infty}\left (\sum_{k=1}^{\infty}\frac{c_k(n)}{2^k}\right )(1-x)^n.$$

So we see $f$ has the desired form.

Claim: $\lim_{x\to 0^+}\dfrac{xf'(x)}{f(x)} = 0.$

Proof: Note that differenitaing termwise in $(1)$ gives a unifomly convergent series on $[a,1]$ for any $a\in (0,1].$ Thus

$$f'(x) = \sum_{k=1}^{\infty}\frac{x^{1/k-1}}{k2^k}$$

for $x\in (0,1].$ Therefore

$$\frac{xf'(x)}{f(x)} = \frac{ \sum_{k=1}^{\infty}x^{1/k}/(k2^k)}{ \sum_{k=1}^{\infty}x^{1/k}/2^k}.$$

Let $N\in \mathbb N.$ Define $S_N(x) = \sum_{k=1}^{N}\dfrac{x^{1/k}}{2^k},$ $T_N(x) = \sum_{k=N+1}^{\infty}\dfrac{x^{1/k}}{2^k}.$ Then

$$\tag 2\frac{xf'(x)}{f(x)} \le \frac{S_N(x) +T_N(x)/N}{S_N(x) + T_N(x)} \le \frac{S_N(x)}{x^{1/(N+1)}/2^{N+1}} + \frac{1}{N}.$$

As $x\to 0^+,$ the right side of $(2) \to 0+1/N = 1/N.$ Because $N$ was arbitrary, the $\limsup_{x\to 0^+}$ of the left side of $(2)$ is $0,$ proving the claim.