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