Answer
$$\frac{1}{2}{e^{ - \frac{1}{{{x^2}}}}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{ - 1/{x^2}}}}}{{{x^3}}}} dx \cr
& {\text{use the property of negative exponents }}\frac{1}{{{x^n}}} = {x^{ - n}} \cr
& \int {{e^{ - {x^{ - 2}}}}\left( {{x^{ - 3}}} \right)} dx \cr
& {\text{set }}u = - {x^{ - 2}}{\text{ then }}\frac{{du}}{{dx}} = 2{x^{ - 3}} \to \frac{{du}}{{2{x^{ - 3}}}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{e^{ - {x^{ - 2}}}}\left( {{x^{ - 3}}} \right)} dx = \int {{e^u}\left( {{x^{ - 3}}} \right)} \left( {\frac{{du}}{{2{x^{ - 3}}}}} \right) \cr
& {\text{cancel common terms}} \cr
& = \int {{e^u}\left( {\frac{{du}}{2}} \right)} \cr
& = \frac{1}{2}\int {{e^u}} du \cr
& {\text{integrating}} \cr
& = \frac{1}{2}{e^u} + C \cr
& {\text{replace }} - {x^{ - 2}}{\text{ for }}u \cr
& = \frac{1}{2}{e^{ - {x^{ - 2}}}} + C \cr
& or \cr
& = \frac{1}{2}{e^{ - \frac{1}{{{x^2}}}}} + C \cr} $$
