There are two solutions: $f(x) = -1$ and $f(x) = x + 1$.
The functional equation is
$$f(x - f(y)) = f(f(x)) - f(y) - 1 \quad (1)$$
1. Take $x = 0$ and $y = f(0)$ into (1), we learn that $z = -f(f(0))$ satisfies $f(z) = -1$.

2. Set $y = z$ into (1), we have
$$f(x + 1) = f(f(x)). \quad (2)$$

3. Therefore, (1) becomes
$$f(x - f(y)) = f(x + 1) - f(y) - 1. \quad (3)$$

4. Now we show that $f$ is linear. By applying (3) with $y = x$ and then (2), we obtain
$$f(x + 1) - f(x) = f(x - f(x)) + 1 = f(f(x - 1 - f(x))) + 1.$$
Since (3) shows $f(x - 1 - f(x)) = f(x) - f(x) - 1 = -1$, we have
$$f(x + 1) = f(x) + A,$$
where $A = f(-1) + 1$ is some constant. Standard induction shows that
$$f(x) = Ax + B,$$
where $B = f(0)$.

5. Back to (2), we have
$$Ax + (A + B) = A^2 x + (AB + B).$$
Take $x = 0$ and $x = 1$, we have $A + B = AB + B$ and $A^2 = A$. The second equation leads to $A = 0$ or $A = 1$. If $A = 1$, we have $B = 1$, so $f(x) = x + 1$. If $A = 0$, $f$ would be a constant, and (1) shows that $f(x) = -1$.