**Question:**  

what can be recommended for calculating $f(x)$ that solves $\frac{f(x)}{f(x)+f(1-x)}\approx g(x)$ for $x\in[0,1]$?


I have tried comparing Taylor series, but they look intimidating and I would appreciate suggestions for better solutions



Approximation with special partitions of unity
I don't know what significance the approximate equality has, so I'll just replace $\approx$ by $=$. Any function $g(x)$ on $[0,1]$ can be split into its even $g\_e(x) = (g(x)+g(1-x))/2$ and odd $g\_o(x) = (g(x)-g(1-x))/2$ parts, $g(x) = g\_e(x) + g\_o(x)$. So the desired equation takes the following form, with an immediate implication:
$$
 \frac{f\_e(x)+f\_o(x)}{2f\_e(x)} = g\_e(x) + g\_o(x) \quad \implies \quad
 \frac{f\_e(x)}{2f\_e(x)} = \frac{1}{2} = g\_e(x) .
$$
So the equation will have no solution at all, unless $g\_e(x) = 1/2$. But if that holds, the odd part of the equation simply states that:
$$
 \frac{f\_o(x)}{2f\_e(x)} = g\_o(x) .
$$
That is, the most general solution is parametrized by a choice of $f\_e(x) \ne 0$, with $f\_o(x) = 2 g\_o(x) f\_e(x)$, or $f(x) = f\_e(x) (1+2g\_o(x)) = 2 f\_e(x) g(x)$.


