Functional equation $f(xy)=f(x)+f(y)$ and differentiability

I want to prove the following claim:

If $f:(0,\infty)\to\mathbb{R}$ satisfying $f(xy)=f(x)+f(y)$, and if $f$ differentiable on $x_0=1$, then $f$ differentiable for all $x_0>0$.

Thank you.

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Let $y=1+h/x$. Then
$$f'(x)=\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim\limits_{h\to 0}\frac{f(xy)-f(x)}{h}=\lim\limits_{h\to 0}\frac{f(y)}{h}=\frac{1}{x}\lim\limits_{h\to 0}\frac{f(1+h/x)}{h/x}=\frac{f'(1)}{x}.$$

Set $g(x) = \exp(f(x))$ and then go here or here. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

It’s a bit hazy, but I’ do as follows:
1). Prove that $f(1+x)/x\to f'(1)$ when $x\to 0$
2). Fix $x>0$, observe that $\frac{f(x+xh)-f(x)}{xh}=\frac{f(1+h)}{xh}$; rewriting it as
$$\frac{f(x+\delta)-f(x)}{\delta}=\frac{1}{x}\frac{f(1+\delta/x)}{(\delta/x)}$$
and conclude by remarking that
$$\frac{f(1+\delta/x)}{(\delta/x)} \xrightarrow[\delta\to 0^+]{}f'(1)$$
and thus $$\frac{f(x+\delta)-f(x)}{\delta}=\frac{f'(1)}{x}$$