If $f(x)$ satisfies $2f (x) = f(xy) + f(x/y)$, find $f(x)$ If $f(x)$ is a continuous and differentiable function which satisfies the function equation If $$2f (x) = f(xy) + f\left(\frac xy\right)\quad \forall x,y \in \mathbb{R}^{+}$$ and $f'(1)=1$ then find $f(x)$
I can see that $f(x)=\ln(x)$ is one function which satisfies all the properties but how can it be proved? I tried using first principle of differentiation but wasn't able to obtain the function. Could someone help me with this?
 A: 
Lets find all $f:\Bbb{R}^+\to\Bbb{R}$ continuous  functions with $f(xy)=2f(x)-f\left(\dfrac{x}y\right).$

When $x=y,$ we have $f(x^2)=2f(x)-f(1).$ Now assume for any $r\in\Bbb{N}\cup\{0\},$ $$f(x^r)=r(f(x)-f(1))+f(1).$$ Now $x\to x^r$ and $y\to x$ gives us $f(x^{r+1})=2f(x^r)-f(x^{r-1}).$ Therefore we can proof this claim by strong induction. Using the fact that $f(x^{-r})=-f(x^r)+2f(1),$ we can extend our result to negative integer $r$ values as $$f(x^{-r})=-r(f(x)-f(1))+f(1)$$ Moreover $x\to x^{1/r}$ gives us $$f(x^{1/r})=\dfrac1r(f(x)-f(1))+f(1).$$ Hence the result establishes for rational (positive and negative) $r$ values. Also due to the continuity, we have the result for any $r\in\Bbb{R}.$
Let $x\to e$ and $r\to\log_e{x},$ then $$f(x)=a\ln x+b\,\,\,\,\,\,\forall x\in\Bbb{R}^+$$ where $a=f(e)-f(1)$ and $b=f(1)$ are some constants.
A: 
Problem:   Find all continuous  functions $f:\Bbb{R}^+\to\Bbb{R}$ which satisfy
  $$ \forall_{x\ y>0}\quad 2\cdot f(x)
     \,\ =\,\ f(x\cdot y)\ +\ f\left(\dfrac{x}y\right).$$

Solution:
Let $\ s\ :=\ x\cdot y\ $ and $\ t\ :=\ \frac xy;\ \ $
i.e. $\ \ x\ =\ \sqrt{s\cdot t}\ $ and $\ y\ =\ \sqrt{\frac st}.\ $
Then the above equation is equivalent to:
$$ f\left(\sqrt{s\cdot t}\right)\,\ =\,\ \frac {f(s)+f(t)}2 $$
Define function
$$ g:=f\circ exp: \mathbb R\rightarrow\mathbb R $$
Then:
$$ \forall_{\sigma\ \tau\in\mathbb R}\quad g\left(\frac{\sigma+\tau}2\right)\ =\ \frac{g(\sigma)+g(\tau)}2 $$
It's known that such $\ g\ $ is a linear function (meaning, a polynomial of degree $1$), i.e. there exist $\ A\ B\in\mathbb R\ $ such that:
$$ g(\xi)\ =\ A\cdot\xi+B $$
hence
$$ f(x)\ =\ (f\circ\exp\circ\log)(x)\ =\ (g\circ\log)(x)\ =\ A\cdot\log(x)+B $$

Solution:   The set of required functions $\ f\ $ is
  $$ \{ A\!\cdot\!\log\, +\, B\ :\,\ A\,\ B\in\mathbb R\} $$

A: Let's differentiate your equation with respect to $y$
$$0=xf^{\prime}\left(xy\right)-\dfrac{x}{y^{2}}f^{\prime}\left(\dfrac{x}{y}\right) \tag{1}$$
This is allowed since $f$ is differentiable. Now set $x=y$ and use $f^{\prime}(1)=1$ to have
$$0=xf^{\prime}\left(x^{2}\right)-\dfrac{1}{x} \tag{2}$$
or if we define $u=x^{2}$
$$f^{\prime}(u)=\dfrac{1}{u} \tag{3}$$
Thus $f\left(u\right)=\ln(u)+C$ for $C\in\mathbb{R}$.
Edit 1: Following the answer by @Bumblebee, we can solve this equation in general by not specifying $C_{1}=f^{\prime}(1)$. In this case, Eq. $(3)$ is in fact
$$f^{\prime}(u)=\dfrac{C_{1}}{u} \tag{4}$$
with a solution
$$f(u)=C_{1}\ln(u)+C_{2} \tag{5}$$
