f(x) = x is a solution, but I'm not certain it is the only solution

Plug in several values for x and y.
(x,y)
(0,0): f(-1) + f(0)^2   = -1
(0,1): f(-1) + f(0)f(1) = -1
(0,2): f(-1) + f(0)f(2) = -1
... so it looks like either f(0)=f(1)=f(2)  OR f(0)=0

(1,1): f(0) + f(1)^2   = 1
(1,2): f(1) + f(1)f(2) = 3
(1,3): f(2) + f(1)f(3) = 5
But if f(0)=f(1)=f(2) then s + s^2 = multiple different values
Therefore f(0) = 0
f(-1) = -1
f(1) = 1
f(2) = 2
f(3) = 3

It looks like:   f(x) = x is a solution

a few other plug-ins

(2,2): f(3) + f(2)^2   = 7
(2,3): f(5) + f(2)f(3) = 11
(-2,2): f(-5) + f(-2)f(2) = -9