Solution
= Solution
Set $x=y=0$ in the functional equation:
$$
f(0)=\frac{2f(0)}{1-f(0)^2}.
$$
The denominator is defined. If $f(0)\ne0$, cancellation would give $f(0)^2=-1$, impossible for a real-valued <function>. Hence
$$
\boxed{f(0)=0}.
$$
= Solution
Set $x=y=0$ in the functional equation:
$$
f(0)=\frac{2f(0)}{1-f(0)^2}.
$$
The denominator is defined. If $f(0)\ne0$, cancellation would give $f(0)^2=-1$, impossible for a real-valued <function>. Hence
$$
\boxed{f(0)=0}.
$$