Answer
$$y' = \frac{1}{{3x{{\left( {\ln x + 1} \right)}^{2/3}}}}$$
Work Step by Step
$$\eqalign{
& y = \root 3 \of {\ln x + 1} \cr
& or \cr
& y = {\left( {\ln x + 1} \right)^{1/3}} \cr
& {\text{find the derivative}} \cr
& y' = \left( {{{\left( {\ln x + 1} \right)}^{1/3}}} \right)' \cr
& {\text{use chain rule}} \cr
& y' = \frac{1}{3}{\left( {\ln x + 1} \right)^{ - 2/3}}\left( {\ln x + 1} \right)' \cr
& {\text{use }}\left( {\ln u} \right)' = \frac{{u'}}{u}, \cr
& y' = \frac{1}{3}{\left( {\ln x + 1} \right)^{ - 2/3}}\left( {\frac{1}{x}} \right) \cr
& {\text{simplifying}} \cr
& y' = \frac{1}{{3x}}{\left( {\ln x + 1} \right)^{ - 2/3}} \cr
& y' = \frac{1}{{3x{{\left( {\ln x + 1} \right)}^{2/3}}}} \cr} $$
