Answer
$$y' = 2x{\left( {{x^2} + 3} \right)^{\ln x - 1}}\ln x + \frac{{{{\left( {{x^2} + 3} \right)}^{\ln x}}\ln \left( {{x^2} + 3} \right)}}{x}$$
Work Step by Step
$$\eqalign{
& y = {\left( {{x^2} + 3} \right)^{\ln x}} \cr
& {\text{take logarithm natural on both sides}} \cr
& \ln y = \ln {\left( {{x^2} + 3} \right)^{\ln x}} \cr
& {\text{logarithm properties}} \cr
& \ln y = \ln x\ln \left( {{x^2} + 3} \right) \cr
& {\text{differentiate}} \cr
& \left( {\ln y} \right)' = \left( {\ln x\ln \left( {{x^2} + 3} \right)} \right)' \cr
& {\text{product rule}} \cr
& \frac{{y'}}{y} = \ln x\left( {\ln \left( {{x^2} + 3} \right)} \right)' + \ln \left( {{x^2} + 3} \right)\left( {\ln x} \right)' \cr
& \frac{{y'}}{y} = \ln x\left( {\frac{{3{x^2} - 2}}{{{x^3} - 2x}}} \right)' + \ln \left( {{x^3} - 2x} \right)\left( {\frac{1}{x}} \right) \cr
& y' = y\left[ {\ln x\left( {\frac{{2x}}{{{x^2} + 3}}} \right) + \ln \left( {{x^2} + 3} \right)\left( {\frac{1}{x}} \right)} \right] \cr
& {\text{replace }}y = {\left( {{x^2} + 3} \right)^{\ln x}} \cr
& y' = {\left( {{x^2} + 3} \right)^{\ln x}}\left[ {\ln x\left( {\frac{{2x}}{{{x^2} + 3}}} \right) + \ln \left( {{x^2} + 3} \right)\left( {\frac{1}{x}} \right)} \right] \cr
& {\text{simplify}} \cr
& y' = 2x{\left( {{x^2} + 3} \right)^{\ln x - 1}}\ln x + \frac{{{{\left( {{x^2} + 3} \right)}^{\ln x}}\ln \left( {{x^2} + 3} \right)}}{x} \cr} $$
