This problem does not require calculus to solve. We cross-multiply and simplify the given equation to get \(yx^2-(y+1)x+y-1=0\). The domain of the function is all real numbers as \(x^2-x+1\) does not have any real solutions, so the discriminant of the equation in \(x\) must always be nonnegative, that is,
$$(y+1)^2-4(y)(y-1) \ge 0 \implies 3y^2-6y-1\le0.$$
The maximum and minimum are simply the roots of the LHS, and their sum is \(-\frac{-6}3=\boxed{-2}\). The values of \(y\) are achievable at the \(x\)-values found in the previous answer.
