Answer
$p^{\prime}(z)=(6z+1)^{1/3}(14z+1)$
Work Step by Step
$p(z)=z\cdot(6z+1)^{4/3}=u(z)\cdot v(z)$
Use the product rule
$\color{blue}{[u(z)\cdot v(z)]^{\prime}=u(z)\cdot v^{\prime}(z)+v(z)\cdot u^{\prime}(z)}$
For $v^{\prime}(z)$, we need the chain rule.
$\left[\begin{array}{llll}
u(z)=z & , & v(z)=(6z+1)^{4/3} & \\
u^{\prime}(z)=1 & & v(z)=[w(z)]^{4/3} & \\
& & v^{\prime}(z)=\frac{4}{3}(6z+1)^{1/3}\cdot w^{\prime}(z) & (\text{chain rule)})\\
& & v^{\prime}(z)=\frac{4}{3}(6z+1)^{1/3}\cdot 6 &
\end{array}\right]$.
$p^{\prime}(z)=u(z)\cdot v^{\prime}(z)+v(z)\cdot u^{\prime}(z)$
$=z\displaystyle \cdot\frac{4}{3}(6z+1)^{1/3}\cdot 6+(6z+1)^{4/3}\cdot 1$
$=8z(6z+1)^{1/3}+(6z+1)^{4/3}$
$=(6z+1)^{1/3}[8z+(6z+1)^{3/3}]$
$=(6z+1)^{1/3}[8z+6z+1]$
$=(6z+1)^{1/3}(14z+1)$
