y = x/x^2+1 = 1/(x+1/x)
分母x+1/x为对勾函数：
x＜-1时x+1/x单调增,y = 1/(x+1/x)单调减；
-1＜x＜0时x+1/x单调减,y = 1/(x+1/x)单调增；
0＜x＜1时x+1/x单调减,y = 1/(x+1/x)单调增；
x＞1时x+1/x单调增,y = 1/(x+1/x)单调减.
x∈R,∴y = 1/(x+1/x)单调减区间（-∞,-1）U（1,+∞）,单调增区间（-1,1）
证明：
（一）x∈（-∞,-1）,令x1＜x2＜-1
f(x2)-f(x1) = x2/(x2^2+1)-x1/(x1^2+1)
= (x1^2x2+x2-x1x2^2-x1)/{(x2^2+1)(x1^2+1)}
= (x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}
∵x1＜x2＜-1
∴x2-x1＞0,1-x1x2＜0
∴f(x2)-f(x1)=(x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}＜0,
∴x∈（-∞,-1）时单调减.
（二）x∈（-1,1）,令-1＜x1＜x2＜1
f(x2)-f(x1) = x2/(x2^2+1)-x1/(x1^2+1)
= (x1^2x2+x2-x1x2^2-x1)/{(x2^2+1)(x1^2+1)}
= (x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}
∵-1＜x1＜x2＜1
∴x2-x1＞0,1-x1x2＞0
∴f(x2)-f(x1)=(x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}＞0,
∴x∈（-1,1）时单调增.
（三）x∈（1,∞1）,令1＜x1＜x2
f(x2)-f(x1) = x2/(x2^2+1)-x1/(x1^2+1)
= (x1^2x2+x2-x1x2^2-x1)/{(x2^2+1)(x1^2+1)}
= (x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}
∵1＜x1＜x2
∴x2-x1＞0,1-x1x2＜0
∴f(x2)-f(x1)=(x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}＜0,
∴x∈（1,+∞）时单调减.