Answer
$$\frac{{dy}}{{dx}} = \frac{{{{\left( {1 + x} \right)}^{1/x}}}}{x}\left[ {\frac{1}{{1 + x}} - \frac{{\ln \left( {1 + x} \right)}}{x}} \right]$$
Work Step by Step
$$\eqalign{
& y = {\left( {1 + x} \right)^{1/x}} \cr
& {\text{Take natural logarithm on both sides}} \cr
& \ln y = \ln {\left( {1 + x} \right)^{1/x}} \cr
& {\text{Using the logarithmic property }}\ln {u^n} = n\ln u \cr
& \ln y = \frac{1}{x}\ln \left( {1 + x} \right) \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {\frac{1}{x}\ln \left( {1 + x} \right)} \right] \cr
& {\text{By the product rule}} \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{1}{x}\frac{d}{{dx}}\left[ {\ln \left( {1 + x} \right)} \right] + \ln \left( {1 + x} \right)\frac{d}{{dx}}\left[ {\frac{1}{x}} \right] \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{x}\left( {\frac{1}{{1 + x}}} \right) + \ln \left( {1 + x} \right)\left( { - \frac{1}{{{x^2}}}} \right) \cr
& {\text{Factor}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \left[ {\frac{1}{{1 + x}} - \frac{{\ln \left( {1 + x} \right)}}{x}} \right]\frac{1}{x} \cr
& {\text{Solving for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{y}{x}\left[ {\frac{1}{{1 + x}} - \frac{{\ln \left( {1 + x} \right)}}{x}} \right] \cr
& {\text{Where }}y = {\left( {1 + x} \right)^{1/x}} \cr
& \frac{{dy}}{{dx}} = \frac{{{{\left( {1 + x} \right)}^{1/x}}}}{x}\left[ {\frac{1}{{1 + x}} - \frac{{\ln \left( {1 + x} \right)}}{x}} \right] \cr} $$
