Answer
$$\ln \left| y \right| + \frac{{{y^2}}}{2} = {e^x} + C$$
Work Step by Step
$$\eqalign{
& \left( {1 + {y^2}} \right)y' = {e^x}y \cr
& \cr
& {\text{write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \left( {1 + {y^2}} \right)\frac{{dy}}{{dx}} = {e^x}y \cr
& \cr
& {\text{separating the variables}} \cr
& \left( {1 + {y^2}} \right)dy = {e^x}ydx \cr
& \left( {\frac{{1 + {y^2}}}{y}} \right)dy = {e^x}dx \cr
& \left( {\frac{1}{y} + y} \right)dy = {e^x}dx \cr
& \cr
& {\text{Integrate both sides of the equation}} \cr
& \int {\left( {\frac{1}{y} + y} \right)} dy = \int {{e^x}} dx \cr
& \ln \left| y \right| + \frac{{{y^2}}}{2} = {e^x} + C \cr} $$
