How can I find $f'(2x)$ given that $xf(x)-yf(y)=(x-y)f(x+y)$? it might seem a little bit elementary. $f$ is defined on $\Bbb R$ and it is differantiable. and is not equal to zero.
if $xf(x)-yf(y)=(x-y)f(x+y)$ then find what is $f'(2x)$ equal to?. $(f'(x),2f'(x),xf'(x),4f'(x)$ or $8f'(x)?)$
from that equation
$\dfrac {f\left( x+y\right) -f\left( y\right) } {x}=\dfrac {f\left( x+y\right) -f\left( x\right) } {y}$
How to continue?

The Identity function satisfies problem conditions and you sure that one of these 5 solutions are correct so the only possible one is $f'(2x)=f'(x)$ :) this is not a proof but answer your question if it is correct problem.
