Does there exist a function $f(x)$ such that $f(x)/f(y)$ is a function of $x+y$? Does there exist a real function $f(x)$ such that $f(x)/f(y)$ is a function of only $x+y$, for all $x,y \in \mathbb{R}$?
 A: Let $a$ and $b$ be any two real numbers. Since in your question, $y$ can be any real number, it follows that $f(y) \neq 0$ for all $y$, since division is permitted.
If $\frac{f(a)}{f(b)}$ depends only upon $a+b$, then since $a+b = \frac{a+b}{2} + \frac{a+b}2$, we have that: $\frac{f(a)}{f(b)} = \frac{f\left(\frac{a+b}2\right)}{f\left(\frac{a+b}2\right)} = 1$, since the expression $\frac{f\left(\frac{a+b}2\right)}{f\left(\frac{a+b}2\right)}$ depends only upon the sum of the operands, which is $a+b$.
Hence, $f(a) = f(b)$ for all $a,b$. It follows that $f$ is a constant function.
EDIT : This proof cannot be used to conclude that $\frac{f(x)}{f(y)}$ cannot non-trivially  depend on $x-y$. 
The reason is simple: If $x+y$ can be fixed, then $x=y$ is possible, which allows us to determine the quotient.
However, if $x-y$ be fixed and non-zero, then $x=y$ is not possible, and nothing about the quotient can be concretely said. 
For example, if $f(x) = e^x$, then $\frac{f(x)}{f(y)} = e^{x-y}$ depends only on $x-y$, a non-trivial solution. This is the crux of my proof.
A: Note that from $f(x)/f(y) = g(x+y)$ follows that:


*

*$f(x) \equiv f(0) \cdot g(x)$ (plugging $y=0$),

*$f(0) \equiv f(y) \cdot g(y)$ (plugging $x=0$).


If we combine these two equalities, we get that
$$ f(x) \equiv \bigl ( f(x) g(x) \bigr ) g(x) $$
or
$$ f(x) \cdot \bigl ( 1 - g(x)^2 \bigr ) \equiv 0. $$
If we want to avoid troubles with division by zero, then $f(x)$ is never zero, and from this follows that 
$$ g(x)^2 - 1 = 0.$$
As @zwim noted, we can't straightforwardly jump to conclusion that $g(x) \equiv 1$ or $g(x) \equiv -1$. However, note that $1 \equiv f(x)/f(x) = g(2x)$. This holds for all $x \in \mathbb{R}$ (since we prohibited $f(x)$ for taking zero value). Thus, $g(x) \equiv 1$. 
So, $f(x) = f(0)$ and constant non-zero functions are the only solutions for your problem.
