Solutions

Let

$$g(x)=\begin{cases}\frac{f(x)}{x},&x\ne0,\\ f'(0),&\text{otherwise.}\end{cases}$$

Then, for $x\ne0$, we have $f(x)=xg(x)$. Also, we have $f(0)=0\cdot f'(0)=0$. Thus, $f(x)=xg(x)$ for all $x$. Lastly, $g$ is continuous at $0$ because

$$g(0)=f'(0)=\lim_{h\to0}\frac{f(h)}{h}=\lim_{h\to0}g(h).$$