'Inverse' type transform of a given functional equation Consider the following functional equation :
$$ \frac{f(a+ix)}{f(a-ix)} = g(x) $$
Here , $a$ is a constant
$f$ and $g$ are smooth functions .

Question : How to write $f(x)$ in terms of function $g$ (assuming we know everything about $g$) ?
i.e is there an 'inverse' type of transform we can apply here ?

 A: Define the function $f^{*}(x)=f(a+ix)$. Then we have
$$\frac{f^*(x)}{f^*(-x)}=g(x)\tag{1}$$
Notice that, by substituting $x\mapsto -x$, we have
$$\frac{f^*(-x)}{f^*(x)}=g(-x)$$
which,together with the first equation, implies $g(x)g(-x)=1$ for all $x$.
There is more than one possible function $f^*$ satisfying this functional equation. A general solution is given by
$$f^*(x)=g^{1/2}(x)\cdot E(x)\tag{2}$$
where $g^{1/2}$ is any function satisfying $(g^{1/2}(x))^2=g(x)$ for all $x$, and $E(x)$ is any even function such that $E(x)=E(-x)$.
To see why this works, notice that
$$\frac{f^*(x)}{f^*(-x)}=\frac{g^{1/2}(x)E(x)}{g^{1/2}(-x)E(-x)}=\frac{g^{1/2}(x)E(x)}{(g^{1/2}(x))^{-1}E(x)}=g(x)$$
Similarly, one can prove that for any solution $f^*$ to your functional equation and square root $g^{1/2}$ of $g$, the function $E=f^*/g^{1/2}$ is in fact even, proving that (2) is actually the general solution to (2) for any square root $g^{1/2}$ of $g$.
As for smoothness - I’m not sure how to deal with that part of the question, so I’ll leave it to you.
