Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $$f\left (x+xy+f(y) \right )=\left (f(x)+ \frac 12 \right )\left (f(y)+ \frac 12 \right ).$$ for every $x,y \in \mathbb R$.
My work so far:
1) $y=-1$: $$f(f(-1))=\left(f(x)+\frac 12 \right ) \cdot \left(f(-1)+\frac 12 \right ).$$ So, if $f(-1) \not = -\frac 12$ that $f=const$ ($c=(c+1/2)^2$ - contradiction)
So, $f(-1)=-\frac 12$.
2) $x=0$: $$f(f(y))=\left(f(0)+\frac 12 \right ) \cdot \left(f(y)+\frac 12 \right ).$$ $y=0$ $$f(x+f(0))=\left(f(0)+\frac 12 \right ) \cdot \left(f(x)+\frac 12 \right )$$
From $f(-1) = -\frac{1}{2}$ we obtain $$ f \left( - \frac{1}{2} \right) = f (f (-1)) = 0.$$ So by choosing $y= - \frac{1}{2}$ we obtain $$f\left( \frac{1}{2} x \right) = f\left(x - \frac{1}{2} x + f\left( - \frac{1}{2} \right) \right) = \frac{1}{2} f(x) + \frac{1}{4}.$$ So $f(0) = \frac{1}{2}$.
Choosing $y=0$ gives $$f\left(x + \frac{1}{2}\right)=f(x) + \frac{1}{2}.$$ Combining this gives for $x= \frac{z}{2^n}$ with $z \in \mathbb{Z}$ and $n \in \mathbb{N}$ that $f(x) = x + \frac{1}{2}$.
Now, if $f$ is continuous, we have $f(x) =x + \frac{1}{2}$ as the Dyadic numbers are dense on the real line.
However, $f$ does not need to be continuous, but for $x= \frac{m}{n}$ with $m,n \neq 0$ and $y=n-1$ we have $$m+n = f \left( x+xy + f(y) \right) = n f(x) + \frac{n}{2}$$ and so $$f \left( \frac{m}{n} \right) = \frac{2m+n}{2n} = \frac{m}{n} + \frac{1}{2}.$$ So for all $q \in \mathbb{Q}$ we have $f(q) = q + \frac{1}{2}$.