Answer
$$m = - \frac{1}{{16}}$$
Work Step by Step
$$\eqalign{
& y = - {x^{ - 3}} + {x^{ - 2}},\,\,\,\,\,\,\,x = 2 \cr
& {\text{find the derivative of the function}} \cr
& \frac{{dy}}{{dx}} = {D_x}\left( { - {x^{ - 3}} + {x^{ - 2}}} \right) \cr
& {\text{use sum rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = {D_x}\left( { - {x^{ - 3}}} \right) + {D_x}\left( {{x^{ - 2}}} \right) \cr
& \frac{{dy}}{{dx}} = - {D_x}\left( {{x^{ - 3}}} \right) + {D_x}\left( {{x^{ - 2}}} \right) \cr
& {\text{use the power rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = - \left( { - 3{x^{ - 4}}} \right) + \left( { - 2{x^{ - 3}}} \right) \cr
& \frac{{dy}}{{dx}} = 3{x^{ - 4}} - 2{x^{ - 3}} \cr
& \frac{{dy}}{{dx}} = \frac{3}{{{x^4}}} - \frac{2}{{{x^3}}} \cr
& {\text{find the slope at }}x = 2{\text{ evaluating the derivative at }}x = 2 \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 2}} = \frac{3}{{{{\left( 2 \right)}^4}}} - \frac{2}{{{{\left( 2 \right)}^3}}} \cr
& {\text{simplifying}} \cr
& m = \frac{3}{{16}} - \frac{2}{8} \cr
& m = \frac{{3 - 4}}{{16}} \cr
& m = - \frac{1}{{16}} \cr} $$
