Suppose $f$ is smooth (up to smooth continuation at the origin) but not analytic, $f(0)=0$, is positive and strictly monotone when $x>0$, and even. Assume $f^{(n)}(0)=0$ for all $n$, do all derivatives of $\log f(x)$ tend to $+\infty$ or $-\infty$ when taking the limit $x\to 0$?
For example, it is easily checked that $e^{-\frac1{x^p}}$ satisfies these criteria for any $p>0$ even.
If not, please provide a valid counter-example.
My solution:
I will show that for $m\ge 0$, $n\ge1$, $$\lim_{x\to 0} |f^{(m+1)}/[f^{(m)}]^n| = \infty.$$ Suppose $n>1$, and rather that the quotient converge to some $k<\infty$, then $|[(f^{(m)})^{-n+1}]'| \le k$ inside some small interval. This contradicts the assumption that that $(f^{(m)})^{-n+1}\to\infty$ when $x\to 0$. The case for $n=1$ is similar using $\log$ instead.
Now expanding derivatives of $\log f$ in terms of derivatives of $f$, this essentially shows $(\log f)^{(m)} \propto f^{(m)}/f^m$ when $x$ is small. In fact the proof implies an improved version of the proposition will hold for finitely differentiable conditions.