For the first problem, one may be tempted to use L'Hopital's Rule:
Noting that $f(x)>0$ for $x$ sufficiently close to $x_0$:
$\ln \biggl[\,\Bigl ({f(x)\over f(x_0)}\Bigr)^{1\over \ln x-\ln x_0}\,\biggr] = {\ln f(x)-\ln f(x_0)\over \ln x-\ln x_0}.$
We have: \tag{1}\lim_{x\rightarrow x_0} {\ln f(x)-\ln f(x_0)\over \ln x-\ln x_0} = \lim_{x\rightarrow x_0} {{f'(x)/ f(x)} \over 1/x} =\lim_{x\rightarrow x_0} { {xf'(x)\over f(x)} }={ {x_0f'(x_0)\over f(x_0)} }.
So \lim_{x\rightarrow x_0}\Bigl ({f(x)\over f(x_0)}\Bigr)^{1\over \ln x-\ln x_0}= \exp\Bigl({ {x_0 f'(x_0)\over f(x_0)} }\Bigr).
But, this line of reasoning is incorrect. We do not know that the last limit appearing in (1) exists.
See the other answers for correct arguments.