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$f(x) + f(y) + f(x)f(y) = 1$

For all values of $x,y \in R$

$f'(x) = ?$

Putting y = x we have,

$2f(x) + {\left[ {f(x)} \right]^2} = 1$

$2f(x) + {\left[ {f(x)} \right]^2} + 1 = 1 + 1$

${[f(x) + 1]^2} = 2$

$f(x) =  - 1 \pm \sqrt 2 $

Since f(x) is a constant function, $f'(x) = 0$.