Show that $f$ is differentiable on $(−1, 1)$. Let $f$ be a function defined on the interval $(−1, 1)$ such that for all $x, y \in (−1, 1)$, $$f(x + y) = \frac{{f(x) + f(y)}}{{1 - f(x)f(y)}}.$$ Suppose that $f$ is differentiable at $x = 0$, show that $f$ is differentiable on $(−1, 1)$.
Need your kind guidance and help. Thanks in advance!
 A: Since $\,f(0)=0\,$, we get:
$$f'(0)=\lim_{x\to 0}\frac{f(x)}{x}\Longrightarrow \,\forall w\in (-1,1)\setminus\{0\}:$$
$$f'(w)=\lim_{h\to 0}\frac{f(w+h)-f(w)}{h}=\lim_{h\to 0}\frac{\frac{f(w)+f(h)}{1-f(w)f(h)}-f(w)}{h}=$$
$$=\lim_{h\to 0}\frac{\left(f(w)^2+1\right)f(h)}{h\left[1-f(w)f(h\right]}=\lim_{h\to 0}\frac{f(h)}{h}\cdot\lim_{h\to 0}\frac{f(w)^2+1}{1-f(w)f(h)}$$
where the last step is justified since both limits on the right hand side exist finitely...
A: Letting $x=y=0$, we obtain $f(0)=\frac{2f(0)}{1-f(0)^2}$, hence $f(0)=0$ or $1-f(0)^2=2$. Since the latter is impossible, $f(0)=0$.
Then we are given that the limit
$$\tag1\lim_{h\to0}\frac{f(h)}h=\lim_{h\to0}\frac{f(h)-f(0)}h=f'(0)$$
exists and of course even more so we have
$$\tag2\lim_{h\to0}f(h)=0.$$
We find
$$\begin{align}f(x+h)-f(x)&=\frac{f(x)+f(h)}{1-f(x)f(h)}-f(x)\\
&=\frac{f(x)+f(h)-f(x)(1-f(x)f(h))}{1-f(x)f(h)}\\
&=f(h)\cdot\frac{1+f(x)^2}{1-f(x)f(h)}\end{align}$$
and thus see (using (1) and (2)) that the derivative at $x$ exists:
$$ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{f(h)}h\cdot \lim_{h\to0}\frac{1+f(x)^2}{1-f(x)f(h)}=f'(0)\cdot (1+f(x)^2).$$
A: Assuming differentiability at $x=0$, we have that
$$\lim\limits_{h\to 0}\frac{f(h)-f(0)}{h}=\ell$$
exists.
Now, note that from $$f(x + y) = \frac{{f(x) + f(y)}}{{1 - f(x)f(y)}}.$$ we get that
$$f(x +h)-f(x) = \frac{{f(x) + f(h)}}{{1 - f(x)f(h)}}-f(x)$$
$$f(x +h)-f(x) = \frac{{f(x) + f(h)}}{{1 - f(x)f(h)}}-\frac{{f(x)({1 - f(x)f(h)})}}{{1 - f(x)f(h)}}$$
$$f(x +h)-f(x) = \frac{{f(x) + f(h)}}{{1 - f(x)f(h)}}-\frac{{{f(x) - f(x)^2f(h)}}}{{1 - f(x)f(h)}}$$
$${f(x +h)-f(x)}= \frac{{f(x) + f(h)}}{{1 - f(x)f(h)}}-\frac{{{f(x) - f(x)^2f(h)}}}{{1 - f(x)f(h)}}$$
$$f(x +h)-f(x) = \frac{{f(x) + f(h)-f(x) + f(x)^2f(h)}}{{1 - f(x)f(h)}}$$
$$ \frac{f(x +h)-f(x)}h = \frac{f(h)}{h}\frac{{ 1 + f(x)^2}}{{1 - f(x)f(h)}}$$
Thus
$$\lim\limits_{h\to0} \frac{f(x +h)-f(x)}h =\lim\limits_{h\to0} \frac{f(h)}{h}\frac{{ 1 + f(x)^2}}{{1 - f(x)f(h)}}$$
All you need now is to show $f(0)=0$. The functional equation gives  
$$f(0) = \frac{{2f(0)}}{{1 - f(0)^2}}.$$
Note that $f(0)^2\neq 1$ since the RHS wouldn't make sense. Thus
$$(1 - f(0)^2)f(0) = 2f(0)$$
$$f(0) - f(0)^3 = 2f(0)$$
$$f(0) + f(0)^3 = 0$$
 $$f(0)(1 + f(0)^2) = 0\implies f(0)=0$$
Thus $f'(0)=\lim\limits_{h\to 0} f(h)/h$, and since $f$ is differentiable at $x=0$ it is continuous there, whence $\lim\limits_{h\to 0} f(h)=f(0)=0$. Finally
$$\lim\limits_{h\to0} \frac{f(x +h)-f(x)}h =\lim\limits_{h\to0} \frac{f(h)}{h}\frac{{ 1 + f(x)^2}}{{1 - f(x)f(h)}}$$
$$f'(x) =f'(0)\frac{{ 1 + f(x)^2}}{{1 - 0}}$$
$$f'(x) =f'(0)\left( 1 + f(x)^2\right)$$
