We have $$f(x) = q(x)(x^2-1) + r(x),$$where $ q(x) $ is the quotient and $ r(x) $ is the remainder. Since $ x^2-1 $ is quadratic, the remainder is at most linear; let us write $ r(x) = ax+b $. Observe that $ x=-1 $ and $ x=1 $ are both zeroes of $ x^2-1 $. Thus $ f(1)=r(1) $ and $ f(-1)=r(-1) $. We can use the given formula for $ f(x) $ to compute $ f(1)=-10 $ and $ f(-1)=16 $. Thus we have the system of equations $$\begin{cases} -10 = a\cdot (1) + b,\\ \phantom{-}16 = a\cdot (-1) + b.\end{cases}$$Adding these equations yields $ 6=2b $ and hence $ b=3 $. Substituting into either equation then yields $ a=-13 $. Therefore, $ r(x) = ax+b = \boxed{-13x+3} $.