If $f(\frac{x+y}{x-y})=\frac{f(x)+f(y)}{f(x)-f(y)}$, which of the following statement is correct If $f(\frac{x+y}{x-y})=\frac{f(x)+f(y)}{f(x)-f(y)}$ for $x \ne y$, $x$ and $y$ are integer. Which of the following statement is correct :
(1) $f(0)=0$
(2) $f(1)=1$
(3) $f(-x)=-f(x)$
(4) $f(-x)=f(x)$
I thinks it's (1),(2), and (3) are the correct statement. But I don't know if I'm doing it in right way or not.
Here's my attempt :
For (1):
$\frac{x+y}{x-y}=0\\
x=-y\\
f(0)=\frac{f(x)+f(-y)}{f(x)-f(-y)} 
\\f(0)=0$ (I'm not sure about this part)
For (2) : 
$\frac{x+y}{x-y}=1\\
x+y=x-y\\
-2y=0\\
y=0\\
f(1)=\frac{f(x)+f(0)}{f(x)-f(0)} 
\\f(1)=1$
For (3) : 
$\frac{x+y}{x-y}=-x\\
x+y=xy-x^2\\ ??$ (I'm stuck from this part)
 A: Let $f : \mathbb{Z} \to \mathbb{R}$ solve the functional equation
$$ \forall x, y \in \mathbb{Z} \ \ \text{s.t.} \ \ \frac{x+y}{x-y} \in \mathbb{Z}, \quad f\left(\frac{x+y}{x-y}\right) = \frac{f(x) + f(y)}{f(x) - f(y)}. \tag{*} $$

Step 1. $f(1) = 1$ and $f(0) = 0$.

For $x \neq 0$, plug $y=0$ to see that $f(1)=\frac{f(x)+f(0)}{f(x)-f(0)}$ holds. If $f(1) \neq 1$, then solving in terms of $f(x)$ gives
$$ f(x) = f(0) \frac{f(1) + 1}{f(1) - 1}, $$
and so, $f$ is constant on $\mathbb{Z}\setminus\{0\}$. This immediately yields a contradiction to $\text{(*)}$, since plugging $x = 2$ and $y = 1$ renders the right-hand side of $\text{(*)}$ undefined. This forces $f(1) = 1$, which then implies $f(0) = 0$.

Step 2. $f(-x) = -f(x)$.

In view of Step 1, it suffices to assume $x \neq 0$. Plugging $y = -x$, we have
$$ 0 = f(0) = \frac{f(x) + f(-x)}{f(x) - f(-x)}, $$
and so, $f(x) + f(-x) = 0$ and the desired claim follows.

By Step 1 and 2, (1), (2) and (3) are correct. Also, notice that $f(x) = x$ solves $\text{(*)}$ but does not satisfy (4). So (4) cannot be true.
A: Your argument for (1) fails; if we choose $x=-y$ so that $\frac{x+y}{x-y}=0$, that doesn't do what you claimed to the right hand side. Instead, we get
$$f(0)=\frac{f(-y)+f(y)}{f(-y)-f(y)}$$
by substituting $-x$ everywhere $y$ shows up.
Not enough information yet. That'll be zero if we can prove (3), but we haven't.
For (2) - well, that'll work if we can prove (1).
For (3), you know you don't have it.
Wait, $x$ and $y$ are restricted to integers in the functional equation? That'll limit us.
So, how do we prove (3)? Well, let's go back to what we got looking into (1);
$$f(0)=\frac{f(-y)+f(y)}{f(-y)-f(y)}$$
for all nonzero integer $y$. But then, it's also true at $-y$, and
$$f(0)=\frac{f(y)+f(-y)}{f(y)-f(-y)}=-\frac{f(-y)+f(y)}{f(-y)-f(y)}=-f(0)$$
Aha - we must have $f(0)=0$, and (1) is true. From there, $f(-y)+f(y)=0$ for all integer $y$, and we have (3) for integer $y$. For other rational values $f(\frac mn)$, we can write $\frac{2m}{2n}=\frac{(m+n)+(m-n)}{(m+n)-(m-n)}$ and
$$f\left(\frac mn\right) = \frac{f(m+n)+f(m-n)}{f(m+n)-f(m-n)}$$
$$f\left(\frac {-m}{n}\right) = \frac{f(-m+n)+f(-m-n)}{f(-m+n)-f(-m-n)}=\frac{-f(m-n)-f(m+n)}{f(m+n)-f(m-n)}=-f\left(\frac mn\right)$$
That's (3) for rational $x$. If the domain of $f$ extends any farther, we can't say anything about it there, and (3) may fail.
So, as long as the domain of $f$ is restricted to rationals, we have (1), (2), and (3). Since $f(1)=1$ and $f(-1)=-1$, we definitely don't have (4).
