This may be a trivial problem and I am sorry if it had been asked before.
Given two known functions $f(x)$ and $g(x)$ I want to find $g(f)$. My first idea was to use the chain rule
$$\frac{\partial g}{\partial f}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial f}$$
where the first derivative on the right hand side is easy, while the second $\partial x/\partial f=(\partial f/\partial x)^{-1}$. This yields $\partial g/\partial f$ as a function of $x$. I presume that I want $\partial g/\partial f$ as a function of $f$, and then integrating to get $g(f)$. Am I on the right track?
Edit 1: These two functions arise from experiments and calculations in thermodynamics. We can approximate $f(x)$ well with the Sigmoid function
$$f(x)=\dfrac{a}{1+\exp({x-x_{0}})}.$$
Experimentally determined $g(x)$ can be approximated sufficiently by the parabola $$g(x)=b(x-x_{0})^2+c.$$
The domain of these functions is the vicinity of $x_{0}$, roughly speaking.
Assuming that $f$ is invertible with inverse $f^{-1}$ we have that $$g(x)=g(f^{-1}(f(x)))=(g\circ f^{-1})(f(x))$$ and hence the function $g\circ f^{-1}$ provides $g$ in terms of $f$. In your example you have $$f^{-1}(x)=x_0+\ln{\left(\frac{a}x-1\right)}$$ and hence we get \begin{align} g(x) &=(g\circ f^{-1})(f(x))\\ &=b\left(x_0+\ln{\left(\frac{a}{f(x)}-1\right)}-x_0\right)^2+c\\ &=b\ln^2{\left(\frac{a}{f(x)}-1\right)}+c\\ \end{align}