Finding functional equation in which $g(1)=0$ and $g'(1)=1$ 
Let a function $g:\mathbb{R^{+}}\rightarrow\mathbb{R}$ is a differentiable function such that $2g(x)=g(xy)+g\bigg(\frac{x}{y}\bigg)\forall x,y\in\mathbb{R^+}$ and $g(1)=0$ and $g'(1)=1$. Then $g(x)$ is 

Try: Differentiate both side with respect to $x$, treating $y$ as a constant 
$$2g'(x)=g'(xy)y-g'\bigg(\frac{x}{y}\bigg)\cdot\frac{1}{y}\cdots \cdots (1)$$
Differentiate both side with respect to $y$, treating $x$ as a constant 
$$0=g'(xy)x-g'\bigg(\frac{x}{y}\bigg)\cdot\frac{x}{y^2}\cdots \cdots (2)$$
Could some help me to solve it, Thanks

Something more general may be proved. 
A continuous function $g:\mathbb{R^{+}}\rightarrow\mathbb{R}$ with $g(1)=0$ satisfies $2g(x)=g(xy)+g\bigg(\frac{x}{y}\bigg)\forall x,y\in\mathbb{R^+}$ iff $g= c \ln$ for some real constant $c$.
Obviously $g=c\ln$ satisfies the functional equation and $g(1)=0$.
On the other hand the equation and $g(1)=0$ with $x=y$ gives 
$g(x^2)=2g(x)$. Now, given $u,v>0$, let $x=\sqrt{uv}$ and $y=\sqrt{u/v}$.
Then $g(uv)=g(x^2)=2g(x)=g(xy)+g(x/y)=g(u)+(v)$. Thus the continuous function $f=g\circ \exp:\mathbb{R}\to\mathbb{R}$ satisfies the Cauchy functional equation $f(x+y)=f(x)+f(y)$. Therefore there ist some $c$ such that $f(x)=c x$ for all $x$. Finally note that $g=f\circ\ln$.
