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