Solution: It is easy to know that when x∈(- 1, 1), there is always f(x)+f(-x)=0. That is, on (-1, 1), the function f(x) is odd function. ∴ According to the property of odd function, the function f(x) has the same monotonicity at (-1, 0) and (0, 1), so we only need to discuss the monotonicity of the function f(x) at (0, 1). When 0 0. And 1/f(x)=x+( 1/x). According to the monotonicity of "tick function", on (0, 1), the function