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