Answer
$$f''\left( x \right) = 20{x^2}{e^{ - {x^2}}} - 10{e^{ - {x^2}}},f''\left( 0 \right) = - {\text{10}},f''\left( 2 \right) = 70{e^{ - 4}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 5{e^{ - {x^2}}} \cr
& {\text{find the derivative of }}f\left( x \right) \cr
& {\text{ }}f'\left( x \right) = \frac{d}{{dx}}\left[ {5{e^{ - {x^2}}}} \right] \cr
& {\text{ }}f'\left( x \right) = 5\frac{d}{{dx}}\left[ {{e^{ - {x^2}}}} \right] \cr
& {\text{use rule }}\frac{d}{{dx}}\left[ {{e^{g\left( x \right)}}} \right] = {e^{g\left( x \right)}}g'\left( x \right).{\text{ set }}g\left( x \right) = - {x^2} \cr
& {\text{ }}f'\left( x \right) = 5{e^{ - {x^2}}}\frac{d}{{dx}}\left[ { - {x^2}} \right] \cr
& {\text{ }}f'\left( x \right) = 5{e^{ - {x^2}}}\left( { - 2x} \right) \cr
& {\text{ }}f'\left( x \right) = - 10x{e^{ - {x^2}}} \cr
& \cr
& {\text{find the derivative of }}f'\left( x \right) \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ { - 10x{e^{ - {x^2}}}} \right] \cr
& {\text{by using the product rule}} \cr
& f''\left( x \right) = - 10x\frac{d}{{dx}}\left[ {{e^{ - {x^2}}}} \right] + {e^{ - {x^2}}}\frac{d}{{dx}}\left[ { - 10x} \right] \cr
& f''\left( x \right) = - 10x\left( { - 2x{e^{ - {x^2}}}} \right) + {e^{ - {x^2}}}\left( { - 10} \right) \cr
& {\text{simplifying}} \cr
& f''\left( x \right) = 20{x^2}{e^{ - {x^2}}} - 10{e^{ - {x^2}}} \cr
& \cr
& {\text{find }}f''\left( 0 \right){\text{ and }}f''\left( 2 \right) \cr
& f''\left( 0 \right) = 20{\left( 0 \right)^2}{e^{ - {{\left( 0 \right)}^2}}} - 10{e^{ - {{\left( 0 \right)}^2}}} = 0 - 10 = - 10 \cr
& f''\left( 2 \right) = 20{\left( 2 \right)^2}{e^{ - {{\left( 2 \right)}^2}}} - 10{e^{ - {{\left( 2 \right)}^2}}} = 80{r^{ - 4}} - 10{e^{ - 4}} = 70{e^{ - 4}} \cr} $$
