Answer
$$\frac{{dy}}{{dx}} = \frac{{{{\left( {{x^2} - 8} \right)}^{1/3}}\sqrt {{x^3} + 1} }}{{{x^6} - 7x + 5}}\left[ {\frac{{2x}}{{3\left( {{x^2} - 8} \right)}} + \frac{{3{x^2}}}{{2\left( {{x^3} + 1} \right)}} - \frac{{6{x^5} - 7}}{{{x^6} - 7x + 5}}} \right]$$
Work Step by Step
$$\eqalign{
& y = \frac{{{{\left( {{x^2} - 8} \right)}^{1/3}}\sqrt {{x^3} + 1} }}{{{x^6} - 7x + 5}} \cr
& {\text{using radical properties}} \cr
& y = \frac{{{{\left( {{x^2} - 8} \right)}^{1/3}}{{\left( {{x^3} + 1} \right)}^{1/2}}}}{{{x^6} - 7x + 5}} \cr
& {\text{taking the natural logarithm on both sides of the equation}} \cr
& \ln y = \ln \left[ {\frac{{{{\left( {{x^2} - 8} \right)}^{1/3}}{{\left( {{x^3} + 1} \right)}^{1/2}}}}{{{x^6} - 7x + 5}}} \right] \cr
& {\text{quotient rule of logarithms}} \cr
& \ln y = \ln {\left( {{x^2} - 8} \right)^{1/3}}{\left( {{x^3} + 1} \right)^{1/2}} - \ln \left( {{x^6} - 7x + 5} \right) \cr
& {\text{product rule of logarithms}} \cr
& \ln y = \ln {\left( {{x^2} - 8} \right)^{1/3}} + \ln {\left( {{x^3} + 1} \right)^{1/2}} - \ln \left( {{x^6} - 7x + 5} \right) \cr
& {\text{power rule of logarithms}} \cr
& \ln y = \frac{1}{3}\ln \left( {{x^2} - 8} \right) + \frac{1}{2}\ln \left( {{x^3} + 1} \right) - \ln \left( {{x^6} - 7x + 5} \right) \cr
& {\text{Differentiate both sides}} \cr
& \ln y = \frac{1}{3}\left( {\frac{{2x}}{{{x^2} - 8}}} \right) + \frac{1}{2}\left( {\frac{{3{x^2}}}{{{x^3} + 1}}} \right) - \frac{{6{x^5} - 7}}{{{x^6} - 7x + 5}} \cr
& {\text{Solve for dy/dx}} \cr
& \frac{{dy}}{{dx}} = y\left[ {\frac{1}{3}\left( {\frac{{2x}}{{{x^2} - 8}}} \right) + \frac{1}{2}\left( {\frac{{3{x^2}}}{{{x^3} + 1}}} \right) - \frac{{6{x^5} - 7}}{{{x^6} - 7x + 5}}} \right] \cr
& {\text{substituting }}y = \frac{{{{\left( {{x^2} - 8} \right)}^{1/3}}\sqrt {{x^3} + 1} }}{{{x^6} - 7x + 5}} \cr
& \frac{{dy}}{{dx}} = \frac{{{{\left( {{x^2} - 8} \right)}^{1/3}}\sqrt {{x^3} + 1} }}{{{x^6} - 7x + 5}}\left[ {\frac{{2x}}{{3\left( {{x^2} - 8} \right)}} + \frac{{3{x^2}}}{{2\left( {{x^3} + 1} \right)}} - \frac{{6{x^5} - 7}}{{{x^6} - 7x + 5}}} \right] \cr} $$
