It is known that f (x) is defined on R and is not equal to 0. For any x, y ∈ R, f (XY) = XF (y) + YF (x) 1. Find the values of F (0), f (1), f (- 1) 2. Judge the parity of F (x) and prove it

Let x = y = 0
Then f (0) = 0 + 0 = 0
Let x = y = 1
Then f (1) = f (1) + F (1)
f(1)=0
Let x = y = - 1
Then f (1) = - f (- 1) - f (- 1)
f(-1)=0
Let y = - 1
f(-x)=0+-f(x)=-f(x)
So it's an odd function

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