Simple Derivation of Functional Equation Question (L'Hospital's Rule)

Question

First, the question is:

$f$ is a differentiable function and $f : R \rightarrow R$

$xf(x)-yf(y)=(x-y)f(x+y)$

$f'(2x)=?$

My approach for problem is using L'Hospital's rule: $$ \frac{xf(x)-yf(y)}{x-y}=f(x+y) $$ Assuming $y=x$ $$ \lim_{y\to{x}} \frac{xf(x)-yf(y)}{x-y} = f(2x)$$ Taking the limit using l'hospital's rule: $$ (yf'(y)+f(y))|_{y=x} = f(2x) $$ Taking the derivative of both sides: $$ f'(x)+xf''(x)+f'(x)=2f'(2x) $$ But the answer is $f'(2x)=f'(x)$

How is this possible?

Answer 1

In equation $$ xf(x)+yf(y)=(x-y)f(x+y), $$ differentiate with respect to $x$ to get $$ xf'(x)+f(x)=(x-y)f'(x+y)+f(x+y). $$ Now, set $y$ to $x$ and to $-x$ to deduce that $$ xf'(x)+f(x)=f(2x)=2xf'(0)+f(0), $$ which proves that $f(x)=xf'(0)+f(0).$