Let \(f:\mathbb R\to\mathbb R\) be a function such that for all real numbers \(x\neq1\), \[f(x-f(x))+f(x)=\frac{x^2-x+1}{x-1}.\] Find all possible values of \(f(2023)\).

[i]Proposed by Linus Tang[/i]
1. **Define the function \( g(x) \):**
   Let \( g(x) = x - f(x) \). Then, the given functional equation can be rewritten in terms of \( g \):
   \[
   f(x - f(x)) + f(x) = \frac{x^2 - x + 1}{x - 1}
   \]
   Substituting \( f(x) = x - g(x) \), we get:
   \[
   f(x - (x - g(x))) + (x - g(x)) = \frac{x^2 - x + 1}{x - 1}
   \]
   Simplifying, we have:
   \[
   f(g(x)) + x - g(x) = \frac{x^2 - x + 1}{x - 1}
   \]
   Since \( f(g(x)) = g(x) \), we get:
   \[
   g(g(x)) = \frac{x^2 - x + 1}{x - 1} - x + g(x)
   \]
   Simplifying further, we obtain:
   \[
   g(g(x)) = \frac{x^2 - x + 1 - x(x - 1) + (x - 1)g(x)}{x - 1}
   \]
   \[
   g(g(x)) = \frac{x^2 - x + 1 - x^2 + x + xg(x) - g(x)}{x - 1}
   \]
   \[
   g(g(x)) = \frac{1 + xg(x) - g(x)}{x - 1}
   \]
   \[
   g(g(x)) = \frac{1 + g(x)(x - 1)}{x - 1}
   \]
   \[
   g(g(x)) = \frac{1}{x - 1}
   \]

2. **Analyze the function \( g(x) \):**
   We need to find \( g \) such that \( g(g(x)) = \frac{1}{x - 1} \). This implies that \( g \) is an involution, meaning \( g(g(x)) = \frac{1}{x - 1} \).

3. **Determine the possible values of \( f(2023) \):**
   We need to find \( f(2023) \). Let \( y = f(2023) \). Then, \( g(2023) = 2023 - y \). We need to ensure that \( g(g(2023)) = \frac{1}{2023 - 1} \):
   \[
   g(2023 - y) = \frac{1}{2022}
   \]
   Since \( g \) is an involution, we have:
   \[
   g(g(2023 - y)) = 2023 - y
   \]
   \[
   g\left(\frac{1}{2022}\right) = 2023 - y
   \]
   Therefore, \( y \neq 0 \) and \( y \neq 2023 + \frac{1}{2022} \).

4. **Verify the conditions:**
   We need to check if all other values are attainable. For \( y \neq 2023 \) and \( y \neq \frac{1}{1 - 2023} \), there exists a function \( g \) satisfying the above condition such that \( g(2023) = y \). Note that \( g(2023) \) cannot equal \( 2023 \) or \( \frac{1}{1 - 2023} \) as it would lead to contradictions.

5. **Conclusion:**
   The possible values of \( f(2023) \) are all \( y \neq 0 \) and \( y \neq 2023 + \frac{1}{2022} \).

The final answer is \( \boxed{ y \neq 0 } \) and \( y \neq 2023 + \frac{1}{2022} \).