Answer
$dy = - \frac{{2x - 3}}{{{{\left( {{x^2} - 3x} \right)}^2}}}dx$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{{x^2} - 3x}} \cr
& {\text{Rewrite the function}} \cr
& y = {\left( {{x^2} - 3x} \right)^{ - 1}} \cr
& {\text{Differentiate the function }} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {{x^2} - 3x} \right)}^{ - 1}}} \right] \cr
& {\text{Use the general power rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = - {\left( {{x^2} - 3x} \right)^{ - 2}}\frac{d}{{dx}}\left[ {{x^2} - 3x} \right] \cr
& \frac{{dy}}{{dx}} = - {\left( {{x^2} - 3x} \right)^{ - 2}}\left( {2x - 3} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{{2x - 3}}{{{{\left( {{x^2} - 3x} \right)}^2}}} \cr
& {\text{Write in the differential form }}dy = f'\left( x \right)dx \cr
& dy = - \frac{{2x - 3}}{{{{\left( {{x^2} - 3x} \right)}^2}}}dx \cr} $$
