Answer
$$y = - \frac{{{e^{1/{x^2}}}}}{{2{x^2}}} + \frac{3}{2}{e^{1/{x^2}}}$$
Work Step by Step
$$\eqalign{
& {x^3}y' + 2y = {e^{1/{x^2}}},{\text{ }}y\left( 1 \right) = e \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& {x^3}\frac{{dy}}{{dx}} + 2y = {e^{1/{x^2}}} \cr
& {\text{Divide both sides by }}{x^3} \cr
& \frac{{dy}}{{dx}} + \frac{2}{{{x^3}}}y = \frac{{{e^{1/{x^2}}}}}{{{x^3}}} \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = \frac{2}{{{x^3}}}{\text{ and }}Q\left( x \right) = \frac{{{e^{1/{x^2}}}}}{{{x^3}}} \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\frac{2}{{{x^3}}}dx} }} = {e^{ - \frac{1}{{{x^2}}}}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{ - \frac{1}{{{x^2}}}}}\left( {\frac{{dy}}{{dx}} + \frac{2}{{{x^3}}}y} \right) = {e^{ - \frac{1}{{{x^2}}}}}\left( {\frac{{{e^{1/{x^2}}}}}{{{x^3}}}} \right) \cr
& {e^{ - \frac{1}{{{x^2}}}}}\frac{{dy}}{{dx}} + \frac{2}{{{x^3}}}{e^{ - \frac{1}{{{x^2}}}}}y = \frac{{{e^0}}}{{{x^3}}} \cr
& {e^{ - \frac{1}{{{x^2}}}}}\frac{{dy}}{{dx}} + \frac{{2{e^{ - \frac{1}{{{x^2}}}}}}}{{{x^3}}}y = \frac{1}{{{x^3}}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {y{e^{ - \frac{1}{{{x^2}}}}}} \right] = \frac{1}{{{x^3}}} \cr
& d\left[ {y{e^{ - \frac{1}{{{x^2}}}}}} \right] = \frac{1}{{{x^3}}}dx \cr
& {\text{Integrate both sides}} \cr
& y{e^{ - \frac{1}{{{x^2}}}}} = \int {\frac{1}{{{x^3}}}} dx \cr
& y{e^{ - \frac{1}{{{x^2}}}}} = - \frac{1}{{2{x^2}}} + C \cr
& {\text{Solve for }}y \cr
& y = - \frac{1}{{2{x^2}{e^{ - \frac{1}{{{x^2}}}}}}} + \frac{C}{{{e^{ - \frac{1}{{{x^2}}}}}}} \cr
& y = - \frac{{{e^{1/{x^2}}}}}{{2{x^2}}} + C{e^{1/{x^2}}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 1 \right) = e \cr
& e = - \frac{{{e^{1/{{\left( 1 \right)}^2}}}}}{{2{{\left( 1 \right)}^2}}} + C{e^{1/{{\left( 1 \right)}^2}}} \cr
& e = - \frac{e}{2} + Ce \cr
& \frac{3}{2}e = Ce \cr
& C = \frac{3}{2} \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = - \frac{{{e^{1/{x^2}}}}}{{2{x^2}}} + \frac{3}{2}{e^{1/{x^2}}} \cr} $$
