First question, let $f,g:(-a,0)\rightarrow(0,\infty)$, introduce the notation:
We say $f(x)=O(g(x))$ in $0$, if there exists constants $c,\epsilon>0$, such that
$$f(x)\leq cg(x) \ \ \forall x\in(-\epsilon,0).$$
Suppose $f,g,f',g':(-a,0)\rightarrow(0,\infty)$, where by $f'$ we denote the derivative of $f$.
And suppose $ f'(x)=O(g'(x))$ in $0$. Then, is it true that:
$$f(x)=O(g(x)), \text{ in } 0 ?$$
Second question: more generally,
if $f,g:(-a,0)\rightarrow\mathbb{R}$, and $f(x)=O(g(x))$ in $0 $ is defined by $$ |f(x)|\leq c|g(x)| \ \ \forall x\in(-\epsilon,0).$$
And suppose
$$ f'(x)=O(g'(x)) \text{ in } 0.$$
Is it true that $$ f(x)=O(g(x)) \text{ in } 0?$$
In some cases I think it's only true that $f(x)+\alpha=O(g(x))$ for some constant $\alpha$.
I can make a bounded function have arbitrarily large derivative. The obvious example is $sin(1/x)$. Note that this is even analytic on the interval you're considering. So smoothness doesn't help. Of course, this function oscillates wildly between positive and negative, but you can imagine that you can make bad counter examples similar to it.
Assuming that all of $f,g,f',g'$ are monotonic is enough, though. The proof uses an alternate phrasing of Big-0 using limits.
If $f'$ is O($g'$), then
$$ \limsup_{x\to 0^-} f'/g' =L $$ for some constant $L\geq 0$. Now, it turns out that by an argument similar to L'hospital, we have
$$ \limsup_{x\to 0^-} f'/g' \geq \limsup_{x\to 0^-} f/g $$
So we conclude $$ \limsup_{x\to 0^-} f/g \leq L $$
As well. The wikipedia page about L'hospital proves this inequality on the way to proving full strength L'hospital.