Answer
$$\frac{{dp}}{{dq}} = - \frac{1}{{2{q^2}}}{\text{ and }}\frac{{{d^2}p}}{{d{q^2}}} = \frac{1}{{{q^3}}}$$
Work Step by Step
$$\eqalign{
& p = \frac{{{q^2} + 3}}{{{{\left( {q - 1} \right)}^3} + {{\left( {q + 1} \right)}^3}}} \cr
& {\text{Simplify the denominator}} \cr
& p = \frac{{{q^2} + 3}}{{{q^3} - 3{q^2} + 3q - 1 + {q^3} + 3{q^2} + 3q + 1}} \cr
& p = \frac{{{q^2} + 3}}{{2{q^3} + 6q}} \cr
& p = \frac{{{q^2} + 3}}{{2q\left( {{q^2} + 3} \right)}} \cr
& p = \frac{{{q^2} + 3}}{{2q\left( {{q^2} + 3} \right)}} \cr
& p = \frac{1}{{2q}} \cr
& p = \frac{1}{2}{q^{ - 1}} \cr
& \cr
& {\text{Find the first derivative}} \cr
& \frac{{dp}}{{dq}} = \frac{d}{{dq}}\left[ {\frac{1}{2}{q^{ - 1}}} \right] \cr
& {\text{use }}\frac{d}{{dq}}\left[ {{q^n}} \right] = n{q^{n - 1}} \cr
& \frac{{dp}}{{dq}} = \frac{1}{2}\left( { - {q^{ - 2}}} \right) \cr
& \frac{{dp}}{{dq}} = - \frac{1}{2}{q^{ - 2}} \cr
& \frac{{dp}}{{dq}} = - \frac{1}{{2{q^2}}} \cr
& \cr
& {\text{Find the second derivative}} \cr
& \frac{{{d^2}p}}{{d{q^2}}} = \frac{d}{{dq}}\left[ { - \frac{1}{2}{q^{ - 2}}} \right] \cr
& {\text{use }}\frac{d}{{dq}}\left[ {{q^n}} \right] = n{q^{n - 1}}{\text{ }} \cr
& \frac{{{d^2}p}}{{d{q^2}}} = - \frac{1}{2}\left( { - 2{q^{ - 3}}} \right) \cr
& \frac{{{d^2}p}}{{d{q^2}}} = {q^{ - 3}} \cr
& \frac{{{d^2}p}}{{d{q^2}}} = \frac{1}{{{q^3}}} \cr} $$
