Answer
$$\frac{{dy}}{{dx}} = \frac{1}{{2x}} + \frac{1}{{3\left( {x + 1} \right)}} - \csc x\sec x$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\frac{{\sqrt x \root 3 \of {x + 1} }}{{\sin x\sec x}}} \right) \cr
& {\text{Using algebraic properties of the natural logarithm function}} \cr
& y = \ln \left( {\sqrt x \root 3 \of {x + 1} } \right) - \ln \left( {\sin x\sec x} \right) \cr
& y = \ln \left( {\sqrt x } \right) + \ln \left( {\root 3 \of {x + 1} } \right) - \ln \left( {\frac{{\sin x}}{{\cos x}}} \right) \cr
& y = \frac{1}{2}\ln x + \frac{1}{3}\ln \left( {x + 1} \right) - \ln \left( {\tan x} \right) \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{1}{x}} \right) + \frac{1}{3}\left( {\frac{1}{{x + 1}}} \right) - \frac{{{{\sec }^2}x}}{{\tan x}} \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2x}} + \frac{1}{{3\left( {x + 1} \right)}} - \csc x\sec x \cr} $$
