\(y=\dfrac{x^2+1}{x+1}\implies y'=\dfrac{2x(x+1)-(x^2+1)}{(x+1)^2}=\dfrac{x^2+2x-1}{(x+1)^2}\)
Any line parallel to \(y=x\) will have the same slope of 1, so you're looking for all \(x\) such that \(y'\) above also evaluates to 1.
\(\dfrac{x^2+2x-1}{(x+1)^2}=1\implies x^2+2x-1=(x+1)^2\)
This assumes \(x
eq-1\), which is of course the case because \(x=-1\) lies outside the function's domain.
\(x^2+2x-1=x^2+2x+1\implies -1=1\)
which is not true. This means no tangent line to \(y=\dfrac{x^2+1}{x+1}\) will ever be parallel to \(y=x\).
