Proof verification of function that satisfies $f(xy)=f(x)+f(y)$ A function $f:(0,\infty)\mapsto R$ satisfies the condition $f(xy)=f(x)+f(y) $ for all $x>0,y>0$. If $f$ is differentiable at $1$, prove that $f$ is differentiable at every $c\in (0,\infty)$ and $f'(c)=f'(1)/c$
My attempt: Consider,
$\lim_{x \to c}\{\frac{f(x)-f(c)}{x-c}\}$ $=$ $\lim_{x \to c}\{\frac{2f(x)-f(xc)}{x-c}\}$
Now take $x=yc$, then,
$\lim_{y \to 1}\{\frac{2f(yc)-f(yc^2)}{(y-1)c}\}$ (*)
Now $2f(yc)=2f(y)+2f(c)$ and $f(yc^2)=f(y)+f(c^2)$
Then (*) becomes,
$\lim_{y \to 1}\{\frac{f(y)+2f(c)-f(c^2)}{(y-1)c}\}$
Now, $f(c^2)=2f(c)$ by definition
This implies, $\lim_{y \to 1}\{\frac{f(y)}{(y-1)c}\}$ $=$ $f'(1)/c$
(since $f'(1)= \lim_{y \to 1}\{\frac{f(y)}{y-1}\}$, as taking y=1 we get $f(x)=f(1)+f(x)$ so, $f(1)=0$
Is this correct? Thanks in advance!
 A: This works perfectly. Here's an alternative approach:
Notice that if we can show that $f(x/y)=f(x)-f(y)$ (do this as an exercise) holds for all $x,y>0$, we can write
\begin{align}
\frac{f(x)-f(c)}{x-c} &= \frac{f\left(\frac{x}{c}\right)}{x-c}\\
&= \frac{1}{c}\cdot\frac{f\left(\frac{x}{c}\right)}{\frac{x}{c}-1}\\
&= \frac{1}{c}\cdot\frac{f\left(\frac{x}{c}\right)-0}{\frac{x}{c}-1}\\
&= \frac{1}{c}\cdot\frac{f\left(\frac{x}{c}\right)-f(1)}{\frac{x}{c}-1}
\end{align}
Notice the similarity between
$$\frac{f\left(\frac{x}{c}\right)-f(1)}{\frac{x}{c}-1}\text{ and the difference quotient }\frac{f(h)-f(1)}{h-1}$$
Since $x/c\to 1$ as $x\to c$, this strongly suggests that $f'(c)$ exists and equals $=f'(1)/c$. More precisely, we expect to have
$$\lim_{x\to c}\frac{f\left(\frac{x}{c}\right)-f(1)}{\frac{x}{c}-1}=f'(1)$$
$$\implies f'(c)=\lim_{x\to c}\left(\frac{1}{c}\cdot\frac{f\left(\frac{x}{c}\right)-f(1)}{\frac{x}{c}-1}\right)=\frac{f'(1)}{c}$$
The limit
$$\lim_{x\to c}\frac{f\left(\frac{x}{c}\right)-f(1)}{\frac{x}{c}-1}=f'(1)$$
should be obvious, but just in case, I left an $(\varepsilon,\delta)$ proof of it at the bottom of my post. Right now, I'd like to draw your attention to a comment you made below your post:

"Yeah it could be logarithm!"

It turns out that having $f'(x)=f'(1)/x$ for every $x$ is enough to conclude that $f$ must be a logarithm. More precisely, it must be a constant multiple of $\ln$. To see this, define the function $g:(0,\infty)\to\mathbb{R}$ by
$$g(x)=f'(1)\ln(x)$$
Since $\frac{d}{dx}\ln(x)=\frac{1}{x}$, we will will have
$$g'(x)=\frac{f'(1)}{x}$$
A simple consequence of the mean value theorem is that if two functions have the same derivative on an open interval, they must differ by a constant on that interval. In our case, we've shown that $f$ and $g$ have the same derivative on $(0,\infty)$, so there's a real number $K$ such that
$$f(x)-g(x)=K\text{ for every }x\in(0,\infty)$$
We can find the exact value of $K$ by evaluating $f(x)-g(x)$ for a convenient choice of $x$. We know that $f(1)=0$ and $g(1)=f'(1)\ln(1)=0$, so we have $K=f(1)-g(1)=0$. It follows that
$$f(x)=f'(1)\ln(x)$$
which is what we were aiming to show.
Proof of the limit: Fix an arbitrary $\varepsilon>0$. We want to show that there's a $\delta>0$ such that for every $x>0$ (this ensures that $x/c$ is positive and consequently that $f(x/c)$ is well-defined),
$$0<|x-c|<\delta\implies\left|\frac{f\left(\frac{x}{c}\right)-f(1)}{\frac{x}{c}-1}-f'(1)\right|<\varepsilon$$
The assumption that $f'(1)$ exists implies that there's a $\delta_1>0$ such that for every $y>0$ (this is the domain of $f$),
$$0<|y-1|<\delta_1\implies\left|\frac{f(y)-f(1)}{y-1}-f'(1)\right|<\varepsilon$$
Since $\lim_{x\to c}x/c = 1$, there's a $\delta>0$ such that for every $x\in\mathbb{R}$,
$$0<|x-c|<\delta\implies\left|\frac{x}{c}-1\right|<\delta_1$$
In fact, since $0<|x-c|$, we can assert the stronger implication
$$0<|x-c|<\delta\implies \color{red}{0<}\left|\frac{x}{c}-1\right|<\delta_1$$
It follows that for every positive $x$, if $0<|x-c|<\delta$, then $0<|x/c-1|<\delta_1$, so $x/c$ satisfies the bound $0<|h-1|<\delta_1$. We deduce that
$$\left|\frac{f\left(\frac{x}{c}\right)-f(1)}{\frac{x}{c}-1}-f'(1)\right|<\varepsilon$$
Since $\varepsilon$ was fixed arbitrarily, the preceding argument can be applied to any positive real number, so
$$\lim_{x\to c}\frac{f\left(\frac{x}{c}\right)-f(1)}{\frac{x}{c}-1}=f'(1)$$
A: Here is an alternative solution to prove that $ f'(c)=\frac{f'(1)}{c}$
Substitute $x\to cx$ and $y\to c$
$$f(c^2x)=f(cx)+f(c)$$
Differentiating both sides with respect to $x$
$$c^2f'(c^2x)=cf'(cx)$$
Substitute $x\to\frac1c$
$$\implies c^2f'(c)=cf'(1)$$
$$\implies f'(c)=\frac{f'(1)}{c}$$
