Solutions

We have that $g(a)\ne 0$. Since $g$ is differentiable it also is continuous and therefore there exists an interval $(a-\delta,a+\delta)$ where $g(a)\ne 0$ for some $\delta>0$. It follows that $\frac{f}{g}$ is differentiable over this interval and the derivative is

$$\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}=0.$$

Consider any point $x$ such that $a<x<a+\delta$. By the mean value theorem we have

$$\frac{\left(\frac{f}{g}\right)(x)-\left(\frac{f}{g}\right)(a)}{x-a}=0 \implies \left(\frac{f}{g}\right)(x)-\left(\frac{f}{g}\right)(a)=0 \implies \left(\frac{f}{g}\right)(x)=0 \implies f(x)=0.$$

Thus, $f(x)=0$ for $x\in(a,a+\delta)$. Applying the same mean value theorem argument above but for $x$ such that $a-\delta<x<a$, we get $f(x)=0$ for $x\in(a-\delta,a]$. Hence, $f(x)=0$ for all $x\in(a-\delta,a+\delta)$.