Let $c:=f(0)>0$. Then we have $$g''(x)+cg(x) \leq 0, \: \forall x \geq 0.$$Further we can apply the same argument as here: https://artofproblemsolving.com/community/c7h591668p11179962

As pointed out in a p.m. (thanks @ayan_mathematics_king), the above argument has a flaw, because we don't know whether $g(x)$ is positive to write $g''(x)+cg(x) \leq 0, \forall x \geq 0$ and reduce the problem to the mentioned one. In fact $g$ oscillates around $0,$ so it takes both positive and negative values. But, we can modify a bit the argument used in that Putnam problem and still apply a similar idea. Consider the function $\displaystyle E(x):= g(x)^2 +\frac{g'(x)^2}{f(x)}.$ We have \begin{align*} E'(x)&=2g(x)g'(x) +\frac{2g'(x)g''(x) f(x) -g'(x)^2 f'(x)}{f(x)^2}\\ &= \frac{2f(x)g'(x) \big(f(x)g(x) +g''(x)\big)-g'(x)^2f'(x)}{f(x)^2} \\ &=-\frac{g'(x)^2f'(x) }{f(x)^2}\le 0 \end{align*}This means $E(x)$ decreases, therefore $E(x)\le E(0),\forall x>0$, hence $$g(x)^2 \le E(0)$$thus $g(x)$ is bounded. There is a small issue here, we don't know $f$ is differentiable. But since $f$ is increasing, it is differentiable almost everywhere, and since $E(x)$ is continuous, it's decreasing. That is, the above argument still holds.

Very nice solution.This energy technique is very useful.