Suppose there exists a continuous function satisfying the functional equation, $ f(g(x))=f((h(x))+ f_1(x)$.
Now the way I like to tackle this kind of problem is assuming $f(x)= \sum a_if(x)$. Where all $a_i$'s are real or complex number that may also depend on the choice of $x$.
This method works very well if all $g(x), h(x)$ and $f_1(x)$ are polynomial functions. This method becomes very tedious if any one of the given functions are given in form of quotients.
Say for example:
$$f(x)+f(\frac{x-1}{x})=1+x$$
How to solve this kind of problem efficiently? Please help.
Let $h(x) = \frac{x-1}x$. Then $h^2(x) = \frac1{1-x}$ and $h^3(x) = x$.
Our relation is that $f(t)+f(h(t)) = 1+ t$. We hence have
$$\left\{ \begin{array}{cccccccl} f(x)&+&f\left(\frac{x-1}x\right)& =& 1& +& x&\quad\quad\quad(1)\\ f\left(\frac{x-1}x\right)& +& f\left(\frac1{1-x}\right)& =& 1& +& \frac{x-1}x&\quad\quad\quad(2)\\ f\left(\frac{1}{1-x}\right)& +& f\left(x\right)& =& 1& +& \frac1{1-x}&\quad\quad\quad(3)\\ \end{array}\right.$$
Via $(1)-(2)+(3)$ we find that
$$2f(x) = 1 + x - \frac{x-1}x +\frac1{1-x},$$
that is,
\begin{align} f(x) &= \frac12 \left(x + \frac1x + \frac1{1-x}\right) \\&= \frac{1+x^2-x^3}{2(1-x)x} \end{align}
I hope it's clear how the technique above can be easily generalized whenever $h^n(x) = x$ for some odd $n\geqslant 1$. We would consider, in the obvious notation, the linear combination $(1) - (2) + (3) - \dots - (n-1) + (n)$.
On the other hand, if $h^n(x) = x$ for even $n$, this imposes a condition on the RHS of the relation $(1+t$ in our case$)$, which might help in identifying when no solution exists, but does not otherwise provide a solution per se. This is a consequence of the fact that the square $n\times n$ matrix
$$ \pmatrix{ 1&1&0&0&\dots&0&0\\ 0&1&1&0&\dots&0&0\\ 0&0&1&1&\dots&0&0\\ 0&0&0&1&\dots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\dots&1&1\\ 1&0&0&0&\dots&0&1\\ }$$
is invertible when $n$ is odd, but not when $n$ is even.