Can someone check this? We claim the answer is all $y$ such that $y \neq 0$ and $y \neq 2023+1/2022$. Let $g(x)=x-f(x)$. Then, the condition rewrites as $g(g(x))=1/(1-x),$ for all $x \neq 1$. Equivalently, we need to prove that for all $y \neq 2023$ and $y \neq 1/(1-2023)$, there is a function $g$ satisfying the above condition such that $g(2023)=y$. Note that $g(2023)$ cannot equal $2023$, as otherwise $1/(1-2023)=g(g(2023))=2023,$ a contradiction. Moreover, it cannot equal $1/(1-2023)$ either, as otherwise $g(-1/2022)=g(g(2023))=-1/2022,$ and so $g(u)=u$ for some $u$, implying that $1/(1-u)=g(g(u))=u,$ hence $u-u^2=1,$ a contradiction. Now, it remains to prove that all other values are attainable. We have two cases: Case 1: $y \neq 2022/2023$. Then, we may firstly check that $g^ 6(x)=x$ for all $x$, and so each $x$ belongs in cycles of length $6$ or $3$. Since $g^3(x)=1/(1-g(x)$ and $y=g(2023) \neq 2022/2023,$ we obtain that $g^3(2023) \neq 2023,$ and so $2023$'s cycle has length $6$. Now, for every $x \neq 2023$ define $g(x)=a_x$ for some $a_x \neq x, a_x \neq 1/(1-x)$ and $g(a_x)=1/(1-x)$. For $2023$ we define $g(2023)=y$ and $g(y)=1/(1-2023)$ It is easy to verify that the initial condition is satisfied, and $g$ is well defined, as all values in the cycles are distinct (as $a_x \neq x, a_x \neq 1/(1-x)$) and so the real numbers are partitioned into distinct and disjoint cycles of length $6$. Case 2: $y=2022/2023$. Then, $2023$'s cycle has length $3$. In this case, the construction above works verbatim (we define $g(2023)=2022/2023$ instead), and the real numbers are partitioned into distinct and disjoint cycles of length $6$ (if $x \neq 2023$) or $3$ (if $x=2023$). Hence, the answer is indeed all $y \neq 0,2023+1/(2022)$.