Background: $x,y,a$ are real numbers. $a>0$. $f(x), g(x,y), h(x,y)$ are continuous real functions. $g,h$ are defined on $[0,1]^2$.
$$h(x,y)=\frac{g(x,y)}{f(x)-f(y)}>0, \text{when} \ x\neq y.$$
Here we will say $h$ is monotonic if $h$ is increasing with $x$ and decreasing with $y$: $h(x+a,y)>h(x,y)$ and $h(x,y+a)<h(x,y)$.
Problem: Suppose $f$ is strictly increasing and $h$ is monotonic. Then, what are some properties for $g$, where $g(x,y)=h(x,y)f(x)-h(x,y)g(x)$?
(In other words, what are some necessary or sufficient conditions on $g$ to ensure the existence of $f$?)
Motivation: The problem seems really simple because you can choose your $f$ to make $h$ monotonic. Intuitively, in order to let $h$ increases with $x$, simply let $g$ to increase with $x$, and choose a $f$ that only "slowly" increases with $x$:
$$\frac{g_x(x,y)}{f'(x)}>1.$$
It seems like, after the division, $h$ is still increasing with $x$.
An example for the argument above is $g(x,y)=x-y$. However, below it will be shown that this example does not work.
Example: Suppose that $g(x)=x-y$. Without loss of generality we assume $f(0)=0$.
Due to monotonicity of $h(x,y)$:
$$\frac{g(x+a,y)}{f(x+a)-f(y)}>\frac{g(x,y)}{f(x)-f(y)}$$
$$\frac{g(x,y+a)}{f(x)-f(y+a)}<\frac{g(x,y)}{f(x)-f(y)}.$$
Then, by rearranging:
$$(x+a)f(x)>xf(x+a)$$
$$(-y-a)(-f(y))<(-y))(-f(y+a))$$
Contradiction! That is, $g(x,y)$ cannot be $x-y$.
Here are some conjectures that might help solving this problem:
Conjecture (1): One necessary property for $g(x,y)$ is $g(x,z)>g(x,y)+g(y,z)$ whenever $x>y>z$.
Conjecture (2): $g(x,y)\neq v(x)-v(y)$.