Find all functions $ f: Bbb {R} to Bbb {R} $ as for all $ x, y, z in Bbb {R} $ , $ f (f (x) + yz) = x + f (y) f (z) $

I've been told to do it by proving $ f $ is injective and surjective. I have proved this way: $ y = z = 0 $, so what $ f (f (x)) = x + f ^ 2 (0) $. For all $ b in Bbb {R} $, $ x + f ^ 2 (0) = b $ has a solution, then $ f (f (x)) = b $ has a solution and follows $ f $ I surjective. For $ f (x) = f (y) $, $ f (f (x)) = f (f (y)) $, so $ x + f ^ 2 (0) = y + f ^ 2 (0) $ , so $ x = y $. C & # 39; $ f $ is injective. But how to deduce $ f $ I do not have the ideal for that.