Factoring x + y

Solution 1:

You have that $$\tag 1 g(0)f(0)=0$$ while $$f(1)g(0)=1$$ $$f(0)g(1)=1$$ The first equations says that either $g(0)=0$ or $f(0)=0$, or both. But this contradicts the last two equations.

ADD Note that what I wrote holds also when $x=-y$, and when $x=0$, $y=\alpha$ a constant, and vice-versa. In fact, the last observation means your function cannot be defined for any value of $x$ or $y$. Indeed, we have that for any $\alpha,\beta\in\Bbb R$

$$f(0)g(\alpha)=\alpha$$ $$f(\beta)g(0)=\beta$$

But beacuse of $(1)$, the last relations are impossible, so we cannot define either $f$ or $g$ for any $x\in\Bbb R$.

Solution 2:

This would mean that $\frac{a+y}{b+y}$ would have to be a constant function of $y$ for all $a,b$. But $\frac{a+y}{b+y}=1-\frac{b-a}{b+y}$ is only constant when $a=b$.