Isn't this a logarithmic function? I saw this problem that year. Although I can't get its expression, the logarithmic function satisfies this expression.

There is something like f (x1+x2) = [f (x1)+f (x2)]/[1-f (x1) f (x2)], and the expression cannot be obtained, but the tangent function satisfies this. (1) the function f(x)(f(x)≠0 defined on r satisfies that for any real number x 1, x2, there is always f (x1+x2) = f (x1) f (x2).

Solution: If x0 exists so that f(x0)=0, then

f(x)=f(x-x0+x0)=f(x-x0)f(x0)=0，

This is in contradiction with "0 < f (x)" when x > 0.

∴f(x)=[f(x/2)]^2>； 0,

Let x1; 0，0 & lt； f(x2-x 1)& lt； 1,

∴f(x2)=f(x 1+x2-x 1)=f(x 1)f(x2-x 1)<； f(x 1)，

∴f(x) is a decreasing function.

(2) The function f (X) defined on R is not constant at 0 and satisfies any real number x 1. x2 has f(x1x2) = x2f (x1)+x1f (x2). Try to judge f (x

Solution: Let x2= 1 and get f (x1) = f (x1)+x1f (1).

∴f( 1)=0，

Let x 1=x2=- 1 and get 0=-2f(- 1).

∴f(- 1)=0，

Let x 1=x, x2=- 1, and get f(-x)=-f(x).

F (x) is odd function.