The equation is as follows
$\left({t+1}\right)^{2/3}\left({{2t}^2-1}\right)^3$
was able to get to the point where the chain rule and product rule are used together which yielded:
$\left({t+1}\right)^\frac23 (12t\left({2t-1}\right)^2 + \left({2t}^2-1\right)^3\left(\frac 23{t(t+1}\right)^\frac 13)$
and im stuck :[
It's $$\left(\left({t+1}^\frac 23\right)\left({{2t}^2-1}^3\right)\right)'=2t^2-1+(t+1)(4t)=6t^2+4t-1.$$ We can use the following: $$(uv)'=u'v+uv'.$$
If you meant $$\left((t+1)^\frac 23(2t^2-1)^3\right)'$$ then $$\left((t+1)^\frac 23(2t^2-1)^3\right)'=\frac{2}{3}(t+1)^{-\frac{1}{3}}(2t^2-1)^3+(t+1)^{\frac{2}{3}}\cdot3(2t^2-1)^2\cdot4t=$$ $$=\frac{2(2t^2-1)^3}{3\sqrt[3]{t+1}}+12t(2t^2-1)^2\sqrt[3]{(t+1)^2}=$$ $$=\frac{2(2t^2-1)^2(2t^2-1+18t(t+1))}{3\sqrt[3]{t+1}}=\frac{2(2t^2-1)^2(20t^2+18t-1)}{3\sqrt[3]{t+1}}$$