When we can express the inverse of sum of two functions for example $f=f_1+f_2$ in terms of inverse of two functions $(f_1^{-1},f_2^{-1})$?
Inverse of sum of two functions in terms of individual inverse functions
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$$f(x)=f_1(x)+f_2(x)$$
The considered addition is a 2-ary algebraic function.
We can generalize your question to 2-ary algebraic functions (or to $n$-ary algebraic functions with $n>1$).
We presuppose $f_1,f_2$ are invertible functions over an open domain. We denote their inverses by $f_1^{-1}$ and $f_2^{-1}$ respectively.
Let's investigate the equation
$$A(f_1(x),f_2(x))=y,\tag{1}$$
where $f_1(x),f_2(x)$ are expressions and $A(f_1(x),f_2(x))$ is an algebraic expression in dependence of both $f_1(x)$ and $f_2(x)$ at the same time. So $A$ is an algebraic function with $A\colon D\subseteq\mathbb{C}^2\to\mathbb{C}$. According to the invariance of dimension, $A$ is not injective and therefore not invertible.
But if the functions $f_1,f_2$ are algebraically dependent, an algebraic function $A_0$ exists so that $f_2(x)=A_0(f_1(x))$. Equation (1) then becomes
$$A(f_1(x),A_0(f_1(x)))=y.$$
Because the composition of algebraic functions is an algebraic function, an algebraic function $A_1$ then exists so that
$$A_1(f_1(x))=y.$$
If $A_1$ is bijective, we get
$$x=f_1^{-1}(A_1^{-1}(y)).$$
The main theorem in [Ritt 1925] that is also proved in [Risch 1979] applies this situation for the Elementary functions.
Because each function is composed of bijective pieces, the problem can be applied to partial inverses.
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There is no easy way to do so. (There is an obvious hard way: invert $f_1^{-1}$ and $f_2^{-1}$ to get $f_1$ and $f_2$, sum these to get $f$ and then invert.)
It can well happen that $f_1$ and $f_2$ are invertible but $f$ is not. Or that neither of $f_1$ and $f_2$ is invertible but $f$ is. And you can find an invertible function $f_1$ and a non-invertible function $f_2$ so that $f$ is invertible — and another pair of such functions $f_1$ and $f_2$ so that $f$ is not invertible.
Counterexamples shouldn't be too hard to find once you know they are there. Can you find them?