Does the existence of $f'(0)$ imply that $\lim_{x\to 0}\frac{f(x)-f(\ln(1+x))}{x-\ln(1+x)}=f'(0)$? Does the existence of $f'(0)$ imply 
      $$\lim_{x\to 0}\frac{f(x)-f(\ln(1+x))}{x-\ln(1+x)}=f'(0).$$
Or some extra conditions for the function $f$ is needed?
If $f$ is $C^2$, it is easy to get the result!
 A: Short answer is that there are counterexamples, even $C^1$, if you do not require some form of double differentiability at $0$.
If $f$ is a differentiable function such that $f'(x)-\frac1{1+x}f'(\ln(1+x))$ is not differentiable at $0$, then $\lim_{x\to 0}\frac{f(x)-f(\ln(1+x))}{x-\ln(1+x)}$ does not exist. Call $u(x)=f(x)-f(\ln(1+x))$ and notice that $u'(x)=f'(x)-\frac1{1+x}f'(\ln(1+x))$, $u(0)=0$ and $u'(0)=0$.
$$\lim_{x\to0}\frac{f(x)-f(\ln(1+x))}{x-\ln (1+x)}=\lim_{x\to0}\frac{\frac{f(x)-f(\ln(1+x))}{x^2}}{\frac{x-\ln (1+x)}{x^2}}$$
Since $\lim_{x\to0}\frac{x-\ln(1+x)}{x^2}=\frac12$, the former limit coincides with $$2\lim_{x\to 0}\frac{u(x)}{x^2}=2\lim_{x\to 0}\frac{u(x)-u'(0)x-u(0)}{x^2}$$
Which is $2u''(0)$. Since $u$ is differentiable and $u'(x)$ is assumed not to be differentiable at $0$, such limit can't exists.
Incidentally, $f(x)=\begin{cases}x^2\sin\frac1x&\text{if }x\ne0\\ 0&\text{if }x=0\end{cases}$, for which $f'(x)=\begin{cases}-\cos\frac1x+2x\sin\frac1x&\text{if }x\ne0\\ 0&\text{if }x=0\end{cases}$ should be an example of such a function, because $$\overbrace{\frac1{1+x}\cos\frac1{\ln(1+x)}-\cos\frac1x}^{\text{graph this part on Wolframalpha}}\underbrace{-2\frac{\ln(1+x)}{1+x}\sin\frac1{\ln(1+x)}+2x\sin\frac1x}_{\text{this part is continuous}}$$ does not appear to be continuous.
However, by virtue of the former remarks, as soon as we find a continuous function $g$ such that $g(x)-\frac1{1+x}g(\ln(1+x))$ is not differentiable at $0$, we have a $C^1$ counterexample in $f(x)=\int_0^x g(t)\,dt$.
Provided that $g(x)=x\sin\frac1x$ looks very much like a valid instance of this, an example can be constructed artificially with bump functions. Specifically, we can use this principle:

Let $x_n$ be a strictly decreasing sequence such that $x_n\to 0$. Let $a_n$ be a convergent sequence. Then, there is a continuous function $g:\Bbb R\to\Bbb R$ such that $g(x_n)=a_n$. It can be further required that $\left.g\right\rvert_{\Bbb R\setminus\{0\}}$ is $C^\infty$ and/or that $g$ has compact support.

It can be applied like this: arrange the set $\left\{\frac1n\,:\,n\in\Bbb N_{>0}\right\}\cup\left\{\ln\left(1+\frac1n\right)\,:\,n\in\Bbb N_{>0}\right\}$ in a strictly decreasing sequence $x_n$. Notice that each $x_n$ belongs to either of the two sets, but never to both, because otherwise there would be two positive integers $h,k$ such that $e=\left(1+\frac1h\right)^k$, which is not the case.
Therefore, the sequence $$a_n=\begin{cases}0&\text{if }\exists k,\ x_n=\ln\left(1+\frac1k\right)\\x_n&\text{if }\frac1{x_n}\text{is an odd integer}\\3x_n&\text{if }\frac1{x_n}\text{is an even integer}\end{cases}$$ is well-defined. Since $x_n\searrow 0$ and $0\le a_n\le 3x_n$, $a_n$ converges.
If we consider $g$ as in the lemma, then $$\limsup_{t\to 0}\frac{g(x)-\frac1{1+x}g(\ln(1+x))}x\ge \limsup_{n\to\infty}\frac{g(1/n)-\frac{1}{1+1/n}g(\ln(1+1/n))}{1/n}=3\\\liminf_{t\to 0}\frac{g(x)-\frac1{1+x}g(\ln(1+x))}x\le \liminf_{n\to\infty}\frac{g(1/n)-\frac{1}{1+1/n}g(\ln(1+1/n))}{1/n}=1$$
So $g(x)-\frac1{1+x}g(\ln(1+x))$ is indeed not differentiable at $0$.
A: $\lim_{x\rightarrow 0}{{f(g(x))-f(g(x_0))}\over{x-x_0}}$
$=\lim_{x\rightarrow x_0}{{f(g(x))-f(g(x_0))}\over{g(x)-g(x_0)}}{{g(x)-g(x_0)}\over{x-x_0}}$.
If $f\circ g$ is differentiable at $x_0$ and $g$ is differentiable at $x_0$ with $g`(0)\neq 0$,
$\lim_{x\rightarrow x_0}{{f(g(x))-f(g(x_0))}\over{g(x)-g(x_0)}}$ exists. Take $g(x)=\ln(1+x)$, $g'(0)=1$ so the limit is $f'(0)$.
A: The limit on the left does not necessarily exist. For instance, define
$$f(x)=\begin{cases}
  x^2,&\mbox{if $x$ is rational,}\\
  -x^2,&\mbox{otherwise.}\\
\end{cases}.$$
If $x$ is rational, then $\ln(1+x)$ is irrational, and we get
$$
  \frac{f(x)-f(\ln(1+x))}{x-\ln(1+x)}=\frac{x^2+\ln(1+x)^2}{x-\ln(1+x)}.
$$
Similarly, when $\ln(1+x)$ is rational, then
$$
  \frac{f(x)-f(\ln(1+x))}{x-\ln(1+x)}=\frac{-x^2-\ln(1+x)^2}{x-\ln(1+x)}.
$$
On the other hand, it is an easy exercise to show that
$$
  \lim_{x\to 0}\frac{x^2+\ln(1+x)^2}{x-\ln(1+x)}=4
  \ne -4=\lim_{x\to 0}\frac{-x^2-\ln(1+x)^2}{x-\ln(1+x)},
$$
so the limit does not exist. However, $f'(0)=0$.
