Let $a,h,b$ be real numbers and $a$ is nonzero.
I want to solve the differential equation :
$$f '(x) = a f(x/3)^2 + h f(x/4) +b$$
In case no closed form solutions exist, I would like asymptotics.
If $b=0,h=0$ I guess all solutions are of the form
$$ c x^d $$
??
Assume $f(x) = c x^d$. Then
$$c d x^{d-1} = a c^2 (x/3)^{2d} = \frac{a c^2}{3^{2d}} x^{2d}$$
So $d-1 = 2d$ hence $d=-1$. $- c = 9 a c^2$, divide by $c$ assuming nonzero :
$-1 = 9 a c$ thus $c= \frac{-1}{9a}$.
edit
Just some thoughts :
I think it matters a lot if $a > 3/2$ or if $a < 3/2$. This is based on taking higher derivatives and using the equation repeatedly. In particular if $h^2 + b^2 $ is small but nonzero.
Also I was thinking that this might be related to a generalization of the Akra-Bazzi method.
https://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method
None of these ideas are formal and solid for now, but just sharing.