Solutions

Pick any $y>0$ and let $g(x)=f(xy)$. Since $f(x)$ is differentiable for $x>0$, then so is $g(x)$ for $x>0$, its derivative is

$$g'(x)=f'(xy)y=\frac{1}{xy}y=\frac{1}{x}.$$

If $x=1$, then the equality $f(xy)=f(x)+f(y)$ trivially holds as $f(1)=0$. Otherwise, either $x>1$ or $x<1$ and in either case we can apply the Cauchy mean value theorem. For $x<1$, we have

$$\begin{align*} \big(f(1)-f(x)\big)g'(c)=\big(g(1)-g(x)\big)f'(c) &\implies -\frac{f(x)}{x} = \frac{f(y)-f(xy)}{x} \\ &\implies f(xy)=f(x)+f(y) \end{align*}$$

for some $c\in(x,1)$. For $x>1$ we get the same final equality to hold by applying the Cauchy mean value theorem.