Answer
$$\frac{{dy}}{{dx}} = 2{e^{ - 2/x}}\left( {1 + x} \right)$$
Work Step by Step
$$\eqalign{
& y = {x^2}{e^{ - 2/x}} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^2}{e^{ - 2/x}}} \right] \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{dx}} = {x^2}\frac{d}{{dx}}\left[ {{e^{ - 2/x}}} \right] + {e^{ - 2/x}}\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& {\text{we can use the formula }}\frac{d}{{dx}}{e^u} = {e^u}\frac{{du}}{{dx}}\cr
&{\text{where }}u{\text{ is any differentiable function of }}x \cr
& \frac{{dy}}{{dx}} = {x^2}{e^{ - 2/x}}\frac{d}{{dx}}\left[ { - \frac{2}{x}} \right] + {e^{ - 2/x}}\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& {\text{then}} \cr
& \frac{{dy}}{{dx}} = {x^2}{e^{ - 2/x}}\left( {\frac{2}{{{x^2}}}} \right) + {e^{ - 2/x}}\left( {2x} \right) \cr
& \frac{{dy}}{{dx}} = 2{e^{ - 2/x}} + 2x{e^{ - 2/x}} \cr
& \frac{{dy}}{{dx}} = 2{e^{ - 2/x}}\left( {1 + x} \right) \cr} $$
