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