2. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,

$$
f(x)+y-f(x) y=f(x+f(y))-f(x f(y))
$$
2. By setting $x=0$ and $y=2$ in the original equation (J), we get $f(f(2))=2$. For $x=z-f(1)$ and $y=1$ in (J), we get $f(g(z))=f(z)-1$ for $g(z)=f(1)(z-f(1))$. Specifically, we have $f(g(g(f(2))))=f(f(2))-2=0$; let $a=g(g(f(2)))$. For $y=a$ in (J), we have $a(f(x)-1)=f(0)$, which would mean that $f$ is constant if $a \neq 0$, but that is impossible. Therefore, $a=0$ and $f(0)=0$. Substituting $x=0$ in (J) now gives $f(f(y))=y$. This, among other things, means that $f$ is a bijection.

By substituting $f(y)$ for $y$ in (J), we now get

$$
f(x)+f(y)-f(x) f(y)=f(x+y)-f(x y)
$$

For $x=y=2$, this equation becomes $2 f(2)=f(2)^{2}$, and since $f(2) \neq 0$ (because $f(f(2))=2 \neq f(0)$), it follows that $f(2)=2$. Further, for $x=y=1$, we get $3 f(1)-f(1)^{2}=2$, but $f(1) \neq 2$, so it must be $f(1)=1$. Now, for $y=1$ in (J'), we have $f(x+1)=f(x)+1$.

By substituting $x+1$ for $x$ in (J'), we get $f(x)-f(x) f(y)=f(x+y)-f(x y + y)$; subtracting equation (J') gives $f(x y + y) = f(x y) + f(y)$, i.e., $f(z + y) = f(z) + f(y)$ for all $z, y \neq 0$. This is also true for $y=0$, so $f(z + y) = f(z) + f(y)$ for all $z, y$. Now (J') becomes $f(x) f(y) = f(x y)$, from which, for $x=y=\sqrt{z}$, we get $f(z) = f(\sqrt{z})^{2} \geqslant 0$ for $z \geqslant 0$, meaning the function $f$ is additive and increasing. Since $f(1)=1$, simple induction gives $f(q)=q$ for all rational $q$, and from there $f(x)=x$ for all $x \in \mathbb{R}$.