jmc

Let $a=f(0)$ and $b=f(f(0))$. Setting $y=x$ in the given equation, we get $$[f(x){]}^2-x^2=b(1)$$for all $x$. In particular, for $x=0$, $a^2=b$.

Setting $y=0$ in the given equation, we get $$f(f(x))=(a-1)f(x)+a(2)$$for all $x$.

Substituting $f(x)$ for $x$ in equation (1), we get $$[f(f(x)){]}^2-[f(x){]}^2=b.$$But from equation (2), $[f(f(x)){]}^2=[(a-1)f(x)+a{]}^2=(a^2-2a+1)[f(x){]}^2+2a(a-1)f(x)+a^2$, so $$(a^2-2a)[f(x){]}^2+2a(a-1)f(x)=af(x)[(a-2)f(x)+2(a-1)]=0$$for all $x$.

If $a\ne 0$, then $$f(x)[(a-2)f(x)+2(a-1)]=0$$for all $x$, so $f(x)$ attains at most two different values. But by equation (1), this cannot be the case.

Hence, $a=0$, then $b=0$, so from equation (1), $$[f(x){]}^2=x^2,$$which means $f(x)=x$ or $f(x)=-x$ for all $x$.

Let $x$ be a value such that $f(x)=x$. Then $f(f(x))=f(x)=x$, so by equation (2), $x=-x$, or $x=0$. Hence, the only value of $x$ such that $f(x)=x$ is $x=0$. Therefore, $f(x)=-x$ for all $x$. It is easy to check that this solution works.

Therefore, the sum of all possible values of $f(1)$ is $\boxed{-1}.$