Answer
$$f'\left( x \right) = {e^x}\left( {\frac{1}{{x\ln 2}} + {{\log }_2}\left| x \right|} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {e^x}{\log _2}\left| x \right| \cr
& {\text{Use logarithmic properties}} \cr
& f\left( x \right) = {e^x}\left( {\frac{{\ln \left| x \right|}}{{\ln 2}}} \right) \cr
& f\left( x \right) = \frac{1}{{\ln 2}}{e^x}\ln \left| x \right| \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \underbrace {\frac{d}{{dx}}\left[ {\frac{1}{{\ln 2}}{e^x}\ln \left| x \right|} \right]}_{{\text{Product rule}}} \cr
& f'\left( x \right) = \frac{1}{{\ln 2}}\left( {{e^x}\frac{d}{{dx}}\left[ {\ln \left| x \right|} \right] + \ln \left| x \right|\frac{d}{{dx}}\left[ {{e^x}} \right]} \right) \cr
& {\text{Compute derivatives}} \cr
& f'\left( x \right) = \frac{1}{{\ln 2}}\left( {{e^x}\left( {\frac{1}{x}} \right) + \ln \left| x \right|\left( {{e^x}} \right)} \right) \cr
& f'\left( x \right) = \frac{1}{{\ln 2}}\left( {\frac{{{e^x}}}{x} + {e^x}\ln \left| x \right|} \right) \cr
& {\text{Simplifying}} \cr
& f'\left( x \right) = {e^x}\left( {\frac{1}{{x\ln 2}} + \frac{{\ln \left| x \right|}}{{\ln 2}}} \right) \cr
& f'\left( x \right) = {e^x}\left( {\frac{1}{{x\ln 2}} + {{\log }_2}\left| x \right|} \right) \cr} $$
