Answer
$$h'\left( x \right) = - {\left( {1 - {x^2}} \right)^{0.5}}{\left( {{x^2} - 3x} \right)^{ - 3}}\left( {4x - 6} \right) - x{\left( {{x^2} - 3x} \right)^{ - 2}}{\left( {1 - {x^2}} \right)^{ - 0.5}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {{x^2} - 3x} \right)^{ - 2}}{\left( {1 - {x^2}} \right)^{0.5}} \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {{{\left( {{x^2} - 3x} \right)}^{ - 2}}{{\left( {1 - {x^2}} \right)}^{0.5}}} \right) \cr
& {\text{By the product rule}} \cr
& f'\left( x \right) = {\left( {1 - {x^2}} \right)^{0.5}}\frac{d}{{dx}}\left( {{{\left( {{x^2} - 3x} \right)}^{ - 2}}} \right) + {\left( {{x^2} - 3x} \right)^{ - 2}}\frac{d}{{dx}}\left( {{{\left( {1 - {x^2}} \right)}^{0.5}}} \right) \cr
& {\text{Use the chain rule and the power rule }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}u' \cr
& f'\left( x \right) = - 2{\left( {1 - {x^2}} \right)^{0.5}}{\left( {{x^2} - 3x} \right)^{ - 3}}\left( {2x - 3} \right) \cr
& {\text{ }} + 0.5{\left( {{x^2} - 3x} \right)^{ - 2}}{\left( {1 - {x^2}} \right)^{ - 0.5}}\left( { - 2x} \right) \cr
& {\text{Simplifying}} \cr
& h'\left( x \right) = - {\left( {1 - {x^2}} \right)^{0.5}}{\left( {{x^2} - 3x} \right)^{ - 3}}\left( {4x - 6} \right) - x{\left( {{x^2} - 3x} \right)^{ - 2}}{\left( {1 - {x^2}} \right)^{ - 0.5}} \cr} $$
