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
$\lim\limits_{h\to0} \frac{f(x +h)-f(x)}h =\lim\limits_{h\to0} \frac{f(h)}{h}\frac{{ 1 + f(x)^2}}{{1 - f(x)f(h)}}$
$f'(x) =f'(0)\frac{{ 1 + f(x)^2}}{{1 - 0}}$
$f'(x) =f'(0)\left( 1 + f(x)^2\right)$