$f:\mathbb{R}\to\mathbb{R}$ such that $\frac {f(x+y)}{f(x-y)}=\frac{f(x)+f(y)}{f(x)-f(y)}$ $\forall$ $x\ne y$. Find all continuous functions $f$.
My progress: Denote the assertion as $P(x,y)$. If $P(x,0)$ gives $f(0)=0$ and $P(x,-x)$ gives $f(x) = -f(-x)$. I don't know how to continue. Of course the additive cauchy functional equation works, I didn't found other solution so I think I have to prove that $f(x+y) = f(x) + f(y)$.
Disclaimer: im 99% sure there's an cleaner solution, but i have been at this for hours now, im just happy to be done. If someone knows an better way PLEASE do let me know
Here we go!
Claim 1: $f$ is an odd function
Proof: Already done by you. You reached $f(x) = -f(-x) \implies -f(x) = f(-x)$.$\blacksquare$
Claim 2: $f$ is injective.
Proof: Suppose there exists some $a,b$ such that $a\neq b$ and $f(a) = f(b)$. Then, it follows:
$\frac{f(a+b)}{f(a-b)} = \frac{f(a) + f(b)}{f(a)-f(b)}$
But since we assumed $f(a)=f(b)$, this is absurd, since it would mean the expression that holds for all real $x\neq y$ is undefined at some distinct real values.
But now we know $f$ is both continuous (by problem statement) and injective (by claim 2). Therefore, $f$ is either strictly increasing or stricly decreasing, as otherwise it can't be both continuous and injective.$\blacksquare$
Claim 3: If $f$ is increasing, it is unbounded.
Suppose not. Let $b$ be the least upper bound of $f$. Then, $f(x)<b$ for all $x$. We know $f$ is strictly increasing (and continuous) so, we can always find some $a$ such that $b-f(a) < \epsilon$ for all $\epsilon>0$. Pick $\delta>0$. Then, $f(a+\delta)>f(a)$. Now, notice that:
$\frac{f(2a+\delta)}{f(\delta)} = \frac{f(a+\delta)+f(a)}{f(a+\delta)-f(a)}$
Notice that, as long as $\epsilon < f(a)$, we have that $f(a+\delta)+f(a) > 2f(a) > f(a) + \epsilon > b$.
$\frac{b}{f(\delta)} > \frac{f(2a+\delta)}{f(\delta)} = \frac{f(a+\delta)+f(a)}{f(a+\delta)-f(a)} > \frac{b}{f(a+\delta)-f(a)}$
Meaning,
$\frac{b}{f(\delta)} > \frac{b}{f(a+\delta)-f(a)}$
Notice $f(\delta)$ and $f(a+\delta)-f(a)$ are positive. So,
$f(\delta) < f(a+\delta) - f(a) \implies f(a+\delta) > f(a) + f(\delta)$
Suppose $f$ is increasing. Then, clearly, $\delta > 0$. And, in fact, any $\delta>0$ will satisfy the definition of $\delta$. So, choose $\delta>a$. This means $b>f(a+\delta) > f(a)+f(\delta) > f(a)+f(a) > b$ (absurd). So, if $f$ is increasing, it can't be bounded above. Further, since it is odd, it can't be bounded at all. $\blacksquare$
Claim 4: If $f$ is decreasing, it is unbounded.
Suppose some $f$ that is decreasing and bounded exists. Clearly, defining $g(x)=-f(x)$ gives us another solution to the original equation. But this solution must be increasing and bounded, from it's definition. But claim 3 says not such $g$ exists. Therefore, no such $f$ exists. $\blacksquare$
So, $f$ is strictly increasing (or strictly decreasing), continuous and unbounded.
Define $g_{x}(y)$ such that $f(x+y) = f(x) + f(g_{x}(y))$. Notice since our function $f$ is bijective, $g_{x}$ is well defined for any $x,y$.
Further, we can say:
$\frac{f(x)+f(y)}{f(x)-f(y)} = \frac{f(x+y)}{f(x-y)} = \frac{f(x) + f(g_{x}(y))}{f(x)-f(g_{x}(y))}$
Meaning, $f(x)^{2} -f(x)f(g_{x}(y)) + f(x)f(y) -f(y)f(g_{x}(y)) = f(x)^{2} + f(x)f(g_{x}(y)) -f(x)f(y) - f(y)f(g_{x}(y)) \implies 2f(x)f(y) = 2f(x)f(g_{x}(y)) \implies f(y)=f(g_{x}(y))$
Since $f$ is injective, $y=g_{x}(y)$. Therefore, $f(x+y) = f(x)+f(g_{x}(y)) = f(x)+f(y)$. Meaning, any solution to our equation must satisfy $f(x+y)=f(x)+f(y)$. We finish by using the well known fact that the only continuous solutions of $f(x+y) = f(x)+f(y)$ are $f(x)=cx$ and $f(x)=0$. Substituting back into the original equation, we see only $f(x)=cx$ works.