$f(f(x - y)) = f(x) f(y) - f(x) + f(y) - xy$ Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
$$f(f(x - y)) = f(x) f(y) - f(x) + f(y) - xy$$for all $x,$ $y.$ Find the sum of all possible values of $f(1).$

I tried plugging in combinations of $0$'s and $1$'s, but it didn't really get me anywhere. I would appreciate any help!!
Thanks!!

Hint: try plugging $x=y$ in the formula.
Full answer: This yields $f(f(0)) = f(x)^2-x^2$. Let $c=f(f(0))$. Then $$f(x)^2 = x^2+c.$$ Therefore
$$c^2=f(f(0))^2 = f(0)^2+c = 0^2+2c=2c.$$
Thus $c$ is either $0$ or $2$.
If $c=0$, then $f(1)^2=1$ and therefore $f(1)$ is either $1$ or $-1$. As the function $f(x)=-x$ is a solution of the original equation with $f(1)=-1$, $-1$ is indeed a possible value for $f(1)$.
However, by looking at $x=1$ and $y=0$, we see that $f(f(1)) = -f(1)$, which implies that $f(1)\ne1$, and $1$ is therefore not a possible value for $f(1)$.
If $c=2$, we have that $f(1)^2=3$. Thus $f(f(1))^2 = 5$. But with $x=1$ and $y=0$ again, we find that $\pm\sqrt{5}=f(f(1))=\pm\sqrt{6}\pm\sqrt{2}\pm\sqrt{3}$, which is never true (see this question).
Therefore the only acceptable value for $f(1)$ is $-1$.
