Answer
$$\ln \left| {\ln y} \right| = \arctan x + \ln 2$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{y\ln y}}{{1 + {x^2}}},\,\,\,\,y\left( 0 \right) = {e^2} \cr
& {\text{Separating variables gives}} \cr
& \frac{{dy}}{{y\ln y}} = \frac{{dx}}{{1 + {x^2}}} \cr
& {\text{Integrating both sides of the equation gives}} \cr
& \int {\frac{{dy}}{{y\ln y}}} = \int {\frac{{dx}}{{1 + {x^2}}}} \cr
& \int {\frac{1}{{\ln y}}\left( {\frac{1}{y}} \right)dy} = \int {\frac{1}{{1 + {x^2}}}} dx \cr
& \ln \left| {\ln y} \right| = \arctan x + C \cr
& {\text{Find }}C{\text{ by using the initial value }}y\left( 0 \right) = {e^2} \cr
& \ln \left| {\ln {e^2}} \right| = \arctan 0 + C \cr
& \ln 2 = C \cr
& {\text{Then}}{\text{,}} \cr
& \cr
& \ln \left| {\ln y} \right| = \arctan x + \ln 2 \cr} $$
