Let f(x)+f(y)=fx1−y2+y1−x2    [f(x) is not identically zero]. Then

Given f(x)+f(y)=x1−y2+y1−x2                (1)Replace y by x. Then 2f(x)=f2x1−x23f(x)=f(x)+2f(x)=f(x)+f2x1−x2=fx1−4x21−x2+2x1−x21−x2=fx2x2−12+2x1−x2=fx2x2−1+2x−2x3=f2x3−x+2x−2x3or fx−2x3+2x−2x3=f(x)or f3x−4x3∴ f(x)=0 or 3f(x)=f3x−4x3Consider 3f(x)=f3x−4x3.Replacing x by - x, we get3f(−x)=f4x3−3x                    (2)Also, from (1), f(x)+f(−x)=f(0)Putting x = y = 0 in (1), we have f(0)=0 or f(x)+f(−x)=0Thus, f(x) is an odd function.Now, from (2), −3f(x)=f4x3−3xor f4x3−3x+3f(x)=0