Let $Y=f (x)$ be an implicit equation satisfying the relation $f (|2^x - 1|)-2f (|2^{-x} -1|)=3x-2$. where $x \in \mathbb{R}.$
If $f_1 (x)$ and $f_2 (x)$ are explicit functions obtained from the equations whose domains are $[0,1)$ and $[0,\infty)$ respectively. Then
A) Range of $f_1 (x)$ and $f_2 (x)$ are ?
B) Discuss the derivability of $f_1 (x)$ and $f_2 (x).$
Partial solution.
Changing $x$ into $-x$ (you say you have had this idea) gives:
$$\tag{1} f (|2^{-x} - 1|)-2f (|2^{x} -1|)=-3x-2.$$
Let $u:=f (|2^{x} - 1|)$ and $v:=f (|2^{-x} - 1|)$.
The initial equation can be written: $u-2v=3x-2$,
The second equation, (1), can be written: $v-2u=-3x-2$.
Solving this system, gives:
$$\tag{2}5u=9x+4 \ \iff \ \ u=f(|2^x-1|)=\dfrac{9x+4}{5}$$
Set $X:=2^x-1$. This is equivalent to $X+1=2^x \ \iff \ x=\ln(X+1).$
Plugging these expressions (of $x$ and $X$) into (2) gives:
$$f(|X|)=\dfrac{9\ln(X+1)+4}{5}$$
From this expression, I'm confident that you can proceed.
Remark: I think, without being certain (the text is not clear), that $f_1(X):=f(X)$ and $f_2(X):=f(-X).$