Separation of inverse function Let's consider $$z=\frac{f(x)+f(y)}{K-f(x)+f(y)}$$ where $K$ is a constant.
Is there any formal method to approximate $z$ as $(g(x)+h(y))$ ?
 A: Here is one method (for simplicity I assume the function is defined over the unit square): 
Approximation problem
Suppose we have a continuous function $z :  [0,1]\times [0,1] \rightarrow\mathbb{R}$.  We want to find functions $g:[0,1]\rightarrow\mathbb{R}$, $h:[0,1]\rightarrow\mathbb{R}$ so that $z(x,y) \approx g(x) + h(y)$.   Define the error as: 
$$ Error = \int_0^1 \int_0^1 (g(x)+h(y)-z(x,y))^2dxdy $$
Since we can always subtract a constant from function $g(x)$ and add it to $h(y)$ without affecting separability or changing the error, without loss of generality we assume that $\int_0^1 g(x)dx = 0$. 
Derivation
For each $x$ and $y$, we can imagine $g(x)$ and $g(y)$ as an optimization variable.  Fix $x \in [0,1]$. Taking derivatives with respect to $g(x)$ gives: 
$$ \frac{\partial Error}{\partial g(x)} = \int_0^1 2(g(x)+h(y)-z(x,y))dy = 0  \quad \forall x \in [0,1] $$
Similarly, 
$$ \frac{\partial Error}{\partial g(y)} = \int_0^1 2(g(x)+h(y)-z(x,y))dx = 0 \quad \forall y \in [0,1] $$
The above equalities reduce to: 
\begin{align}
&g(x) + \int_0^1 h(y)dy - \int_0^1 z(x,y)dy = 0 \quad \forall x \in [0,1] \\
&h(y) + \int_0^1 g(x)dx - \int_0^1 z(x,y)dx = 0 \quad \forall y \in [0,1] 
\end{align}
Define $c = \int_0^1 h(y)dy$ and recall that 
$\int_0^1 g(x)dx =0$.  The value $c$ is not yet known since the function $h(y)$ is not yet determined. However, the above two equalities imply: 
\begin{align}
&g(x) = \int_0^1 z(x,y)dy - c \quad \forall x \in [0,1] \\
&h(y) = \int_0^1 z(x,y)dx  \quad \forall y \in [0,1] 
\end{align}
Integrating both sides of the above two equalities gives: 
\begin{align} 
&0 = \int_0^1\int_0^1 z(x,y)dxdy - c \\
&c = \int_0^1 \int_0^1 z(x,y)dxdy 
\end{align}
These last two equations are the same.  So we need:
$$ c = \int_0^1\int_0^1 z(x,y)dxdy $$
We now have our solution: 
Solution:
Define $c = \int_0^1\int_0^1 z(x,y)dxdy$.  The optimal separable functions $g^*$ and $h^*$ are: 
\begin{align}
g^*(x) &= \int_0^1z(x,y)dy - c \quad \forall x \in [0,1]\\
h^*(y) &= \int_0^1 z(x,y)dx \quad \forall y \in [0,1]
\end{align}

The above derivation involves an "informal" differentiation with respect to each variable $g(x)$ and $h(y)$. A formal argument is as follows: Let $g(x)$ and $h(y)$ be continuous functions defined over the unit interval such that $\int_0^1 g(x)dx=0$. We want to show these give an approximation error greater than or equal to that of the above functions $g^*(x), h^*(y)$.  We have: 
\begin{align}
Error &= \int_0^1 \int_0^1 (g(x)+h(y)-z(x,y))^2 dxdy\\
&= \int_0^1\int_0^1 [(g(x)-g^*(x)+h(y)-h^*(y))  + (g^*(x)+h^*(y)-z(x,y))]^2dxdy\\
&=\int_0^1\int_0^1 (g(x)-g^*(x))^2dxdy\\
&+ \int_0^1\int_0^1 (h(y)-h^*(y))^2dxdy\\
&+\underbrace{2\int_0^1\int_0^1 (g(x)-g^*(x))(h(y)-h^*(y))dxdy}_{0}\\
&+\underbrace{2\int_0^1\int_0^1(g(x)+h(y)-g^*(x)-h^*(y))(g^*(x)+h^*(y)-z(x,y))dxdy}_{0}\\
&+\int_0^1\int_0^1(g^*(x)+h^*(y)-z(x,y))^2dxdy \\
&\geq \int_0^1\int_0^1(g^*(x)+h^*(y)-z(x,y))^2dxdy 
\end{align}
where the first underbrace uses $\int_0^1 g^*(x)dx=\int_0^1 g(x)dx=0$, while the second underbrace uses these together with $g^*(x)=\int_0^1z(x,y)dy - c$ and $h^*(y) = \int_0^1 z(x,y)dx$. $\Box$ 
