Answer
$$U'(y)=\frac{10y(y^4+2y^2-1)(y^4+1)^4}{(y^2+1)^6}$$
Work Step by Step
$$U'(y)=\left(\left(\frac{y^4+1}{y
^2+1}\right)^5\right)'=5\left(\frac{y^4+1}{y^2+1}\right)^4\left(\frac{y^4+1}{y^2+1}\right)'=5\left(\frac{y^4+1}{y^2+1}\right)^4\frac{(y^4+1)'(y^2+1)-(y^4+1)(y^2+1)'}{(y^2+1)^2}=
5\left(\frac{y^4+1}{y^2+1}\right)^4\frac{4y^3(y^2+1)-(y^4+1)\cdot2y}{(y^2+1)^2}=5\left(\frac{y^4+1}{y^2+1}\right)^4\frac{4y^5+4y^3-2y^5-2y}{(y^2+1)^2}=5\left(\frac{y^4+1}{y^2+1}\right)^4\frac{2y^5+4y^3-2y}{(y^2+1)^2}=\frac{10y(y^4+2y^2-1)(y^4+1)^4}{(y^2+1)^6}$$