Is it possible to find nonzero solution of functional equation
$\displaystyle
(1+4x)f(2x)=2f(x+2x^2)
$
defined and analitic in some neighbourhood of $\displaystyle 0$
After taking first derivate in FE, by shain rule, we have relation
$\displaystyle 4f(2x)+2(1+4x)[f'(2x)-f'(x+2x^2)]=0$ or $\displaystyle 4f(2x)+2(1+4x)[g(2x)-g(x+2x^2)]=0$, by your notion.
I can't see how you derived $\displaystyle g(2x)=g(x+2x^2)$ from this.