It is known that f (x) is a function defined on R which is not always zero, and f (x) satisfies f (XY) = YF (x) + XF (y) for any x, y in the domain of definition (1) Find the value of F (1), f (- 1) (2) Judge the parity of F (x) white and explain the reason

It is known that f (x) is a function defined on R which is not always zero, and f (x) satisfies f (XY) = YF (x) + XF (y) for any x, y in the domain of definition (1) Find the value of F (1), f (- 1) (2) Judge the parity of F (x) white and explain the reason

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1: Let x = 1, y = 1, f (XY) = f (1) = f (1) + F (1), then f (1) = 0;
Let x = - 1, y = - 1, f (XY) = f (1) = - f (- 1) + - f (- 1) = 0, then f (- 1) = 0;
2. Let x = 0, y = 0, f (XY) = f (0) = 0,
Let y = - 1, f (XY) = f (- x) = - f (x) + XF (- 1) = - f (x), that is, f (- x) = - f (x), and f (x) is a function defined on R which is not always zero, so f (x) is an odd function