Answer
$$h'(t)=\frac{2(2t^2-1)^2(20t^2+18t-1)}{3(t+1)^{1/3}}$$
Work Step by Step
$$h'(t)=((t+1)^{2/3}(2t^2-1)^3)'=
((t+1)^{2/3})'(2t^2-1)^3+(t+1)^{2/3}((2t^2-1)^3)'=\frac{2}{3}(t+1)^{2/3-1}(t+1)'(2t^2-1)^3+(t+1)^{2/3}\cdot3(2t^2-1)^2(2t^2-1)'=
\frac{2}{3}\frac{1}{(t+1)^{1/3}}\cdot1(2t^2-1)^3+3(t+1)^{2/3}(2t^2-1)^2\cdot4t=
\frac{2(2t^2-1)^3+3\cdot12t(t+1)(2t^2-1)^2}{3(t+1)^{1/3}}=
\frac{2(2t^2-1)^2(2t^2-1+18t(t+1))}{3(t+1)^3}=
\frac{2(2t^2-1)^2(20t^2+18t-1)}{3(t+1)^{1/3}}$$
