Answer
$$\frac{{dw}}{{dz}} = - \frac{1}{{{z^2}}} - 1{\text{ and }}\frac{{{d^2}w}}{{d{z^2}}} = \frac{2}{{{z^3}}}$$
Work Step by Step
$$\eqalign{
& w = \left( {\frac{{1 + 3z}}{{3z}}} \right)\left( {3 - z} \right) \cr
& {\text{Use the distributive property}} \cr
& w = \frac{{\left( {1 + 3z} \right)\left( {3 - z} \right)}}{{3z}} \cr
& w = \frac{{\left( {3 - z} \right) + 3z\left( {3 - z} \right)}}{{3z}} \cr
& w = \frac{{3 - z + 9z - 3{z^2}}}{{3z}} \cr
& w = \frac{{3 + 8z - 3{z^2}}}{{3z}} \cr
& w = \frac{3}{{3z}} + \frac{{8z}}{{3z}} - \frac{{3{z^2}}}{{3z}} \cr
& w = \frac{1}{z} + \frac{8}{3} - z \cr
& w = {z^{ - 1}} + \frac{8}{3} - z \cr
& \cr
& {\text{Find the first derivative}} \cr
& \frac{{dw}}{{dz}} = \frac{d}{{dz}}\left[ {{z^{ - 1}} + \frac{8}{3} - z} \right] \cr
& {\text{use }}\frac{d}{{dz}}\left[ {{z^n}} \right] = n{z^{n - 1}}{\text{ and }}\frac{d}{{dz}}\left[ c \right] = 0,{\text{ }}c{\text{ is a constant}} \cr
& \frac{{dw}}{{dz}} = - {z^{ - 2}} + 0 - 1 \cr
& \frac{{dw}}{{dz}} = - {z^{ - 2}} - 1 \cr
& \frac{{dw}}{{dz}} = - \frac{1}{{{z^2}}} - 1 \cr
& \cr
& {\text{Find the second derivative}} \cr
& \frac{{{d^2}w}}{{d{z^2}}} = \frac{d}{{dz}}\left[ { - {z^{ - 2}} - 1} \right] \cr
& {\text{use }}\frac{d}{{dz}}\left[ {{z^n}} \right] = n{z^{n - 1}}{\text{ and }}\frac{d}{{dz}}\left[ c \right] = 0,{\text{ }}c{\text{ is a constant}} \cr
& \frac{{{d^2}w}}{{d{z^2}}} = - \left( { - 2{z^{ - 3}}} \right) - 0 \cr
& \frac{{{d^2}w}}{{d{z^2}}} = 2{z^{ - 3}} \cr
& \frac{{{d^2}w}}{{d{z^2}}} = \frac{2}{{{z^3}}} \cr} $$
