A claim:
If $f:\mathbb{R}^{2}\to \mathbb{R}$ is a continuous function, $f(x,y)$, strictly increasing in each of its arguments, with $f(x(y),y)=0$ for a unique $x(y)$ for each $y$, then $f$ must be a transformation of the function $x-x(y)$.
I know this is a stretch, and probably not true, but I need a result similar to this to solve another related problem here.
Truthfully, I am having a difficult time picturing what the condition "$f(x(y),y)=0$ for a unique $x(y)$ for each $y$" means for the function $f$.
Are there any implications of this condition for the function $f$ as defined? Is my claim true?
Since we already know that $f$ is strictly increasing in $x$, for a fixed $y$, there can be at most one $x$ for which $f(x, y)$ takes on any given value. So all that the $f(x(y), y) = 0$ condition adds is a requirement for all $y$, there exists an $x$ for which $f(x, y) = 0$.
Now consider $f(x, y) = x^3 + y$. This satisfies both conditions, so your claim is that $f(x, y) = F(x + \sqrt[3]y)$ for some $F$. But letting $y = 0$ gives $F(x) = x^3$ for all $x$. So we would have $x^3 + y = \left (x + \sqrt[3]y \right)^3$, which is false.