Find all pairs of functions $f,g: \mathbb R \rightarrow \mathbb R$ such that
1). $f\left(xy(y+1)\right)+y=xf(y)+f\left(x+g(y)\right)$ for all real $x,y$;
2). $f(0)+g(0)=0.$
My work so far:
Case i) $f(0)=g(0)=0$
1) $y=-1; x=z-g(-1) \Rightarrow f(z)=-1-\left(z-g(-1)\right)f(-1)$. Then $f(z)=az+b$. If $f(0)=0$ then $f(z)=z$.
So $$axg(y+1)+y=axy+ax+ag(y)$$ $a\not=0$.
$x=0 \Rightarrow g(y)=\frac{y}{a}$
Hence $$x(y+1)+y=axy+ax+y$$ Then $a=1$, $$f(x)=g(x)=x$$
Case ii) $f(0)\not = 0$
I need help here
Addition:
Maybe a typo in the book, and the problem looks like this:
Find all pairs of functions $f,g: \mathbb R \rightarrow \mathbb R$ such that
1). $f\left(x(y+1)\right)+y=xf(y)+f\left(x+g(y)\right)$ for all real $x,y$;
2). $f(0)+g(0)=0.$
I need help here too
There is no solution possible. To see this, putting $y=0$ in (1) gives $$f(0) = xf(0) + f(x+g(0)) \tag{*}\label{a}$$. Now putting $x=f(0)$ in the above equation gives $$f(0) = (f(0))^2 + f(f(0)+g(0))$$ $$\implies f(0) = (f(0))^2 + f(0)$$ $$\implies f(0)=0$$ This also means $g(0)=0$. Now substituting values of $f(0)$ and $g(0)$ in the equation \eqref{a}, we get $0 = 0 + f(x+0)$, that is $f(x)=0 \forall x\in \mathbb{R}$. But substituting this in equation (1) gives $y=0$ which is clearly a contradiction.(This answer is incorrect since the question was later edited.)
EDIT:
Note that in your proof of $f(z) = az + b$ the fact that $f(0)=0$ has not been used. So this result is true anyway, which almost completes your solution!!. Putting $x=0$ in (1) gives $$f(0) + y = f(g(y))$$ $$\therefore b + y = ag(y) + b$$ $$\therefore y = ag(y)$$. Since $y$ is arbitrary, $a\neq 0$, and hence, $g(y) = \frac{y}{a}$ which gives $g(0) = 0$.