f(x^2 - 1) = lg[ (x^2) / (x^2 - 2) ]
= lg[ ( (x^2-1) + 1) / ( (x^2-1) - 1) ]
令u = x^2 - 1,则：
f(u) = lg[ (u+1) / (u-1) ]
所以：
f(x) = lg[ (x+1) / (x-1) ]
令y = (x+1)/(x-1),则 x = (y+1) / (y-1)
代入上式,得：
f( (y+1) / (y-1) ) = lgy
所以：
f( (x+1) / (x-1) ) = lgx = f(g(x))
所以： g(x) = (x+1) / (x-1)