答：
f [(1-x)/(1+x) ]=(1-x²)/(1+x²)
设t=(1-x)/(1+x)
则t+tx=1-x
(t+1)x=1-t
x=(1-t)/(1+t)
代入得：
f(t)=[1-(1-t)²/(1+t)²] / [1+(1-t)²/(1+t)²]
f(t)=[(1+t)²-(1-t)²] / [(1+t)²+(1-t²)]
=4t / (2+2t²)
=2t/(1+t²)
所以：
f(x)=2x/(1+x²),x≠-1