\(y=\frac{t^2 + 2}{t^4 - 3t^2 + 1}
\\
\\y'=(\frac{t^2 + 2}{t^4 - 3t^2 + 1})'= \frac{(t^2+2)'(t^4 - 3t^2 + 1)-(t^2+2)(t^4 - 3t^2 + 1)'}{(t^4 - 3t^2 + 1)^2} =
\\
\\=\frac{2t(t^4 - 3t^2 + 1)-(t^2+2)(4t^3 - 6t)}{(t^4 - 3t^2 + 1)^2} =\frac{2t^5-6t^3+2t-4t^5+6t^3-8t^3+12t}{(t^4 - 3t^2 + 1)^2} =
\\
\\=\frac{-2t^5-8t^3+14t}{(t^4 - 3t^2 + 1)^2}\)