解由f（x）=x/(x^2-1)
设x1.x2属于（-1,1）且x1＜x2
即f(x1)-f(x2)
=x1/(x1^2-1)-x2/(x2^2-1)
=[x1(x2^2-1)-x2(x1^2-1)]/(x2^2-1)(x1^2-1)
=[x1x2^2-x2x1^2+x2-x1]/(x2^2-1)(x1^2-1)
=[x1x2(x2-x1)+(x2-x1)]/(x2^2-1)(x1^2-1)
=[(x1x2+1)(x2-x1)]/(x2^2-1)(x1^2-1)
由x1.x2属于（-1,1）
知x1x2＞-1
即x1x2+1＞0
又由x1＜x2
即x2-x1＞0
又由x1.x2属于（-1,1）
知(x2^2-1)＜0,(x1^2-1)＜0
即(x2^2-1)(x1^2-1)＞0
即[(x1x2+1)(x2-x1)]/(x2^2-1)(x1^2-1)＞0
即f（x1）-f（x2）＞0
即证明f（x）在（-1,1）是减函数.