Defined function $f:\mathbb{R} \to \mathbb{R}$ which satisfies $$f(2xy) + f(f(x + y)) + f(x + y) = xf(y) + yf(x)$$ for any real numbers $x$ and $y$. Determine the sum of all values $|f(48)|$ which possible.
My working: $$f(2xy) + f(f(x + y)) + f(x + y) = xf(y) + yf(x)$$ Assume: $y=-x$ $$f(2x^2) + f(f(0)) + f(0) = xf(-x) -xf(x)$$ $$2f(x^2) + f(f(0)) + f(0) = -xf(x) -xf(x)$$ $$2f(x^2) + f(f(0)) + f(0) = -2xf(x)$$ $$2f(x^2) + f(f(0)) + f(0) = -2f(x^2)$$ $$f(f(0)) + f(0) = -2f(x^2)-2f(x^2)$$ $$f(f(0)) + f(0) = -4f(x^2)$$
so, I still have doubts about my work with the process. how?
Let $a=f(0)$ and plug $y=0$ in the starting equation. We get $$f(f(x)) + f(x) = ax-a$$ for all $x$ and thus with replacing $x$ with $x+y$ in last equation we get $$f(2xy) + a(x+y)-a = xf(y) + yf(x)$$
Now let $b= f(1/2)$ and put $y=1/2$ and we get $$2f(x) +a(2x-1) = 2xb +f(x)$$ so $f$ is some linear function and now should be easy to finish the task.