Answer
$$f\left( x \right) = - \frac{2}{{{{\left( {x + 1} \right)}^3}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{2{x^2} + 4x + 3}}{{{x^2} + 2x + 1}} \cr
& {\text{Write the function as}} \cr
& f\left( x \right) = \frac{{2{x^2} + 4x + 3}}{{{{\left( {x + 1} \right)}^2}}} \cr
& f\left( x \right) = \left( {2{x^2} + 4x + 3} \right){\left( {x + 1} \right)^{ - 2}} \cr
& {\text{Differentiate by using the product rule}} \cr
& f\left( x \right) = \left( {2{x^2} + 4x + 3} \right)\frac{d}{{dx}}\left[ {{{\left( {x + 1} \right)}^{ - 2}}} \right] + {\left( {x + 1} \right)^{ - 2}}\frac{d}{{dx}}\left[ {2{x^2} + 4x + 3} \right] \cr
& \cr
& {\text{By the formula in the exercice 75}}{\text{. }} \cr
& f\left( x \right) = {\left( {mx + b} \right)^n}\,\, \Rightarrow \,\,\,f'\left( x \right) = nm{\left( {mx + b} \right)^{n - 1}} \cr
& {\text{We obtain}} \cr
& f\left( x \right) = \left( {2{x^2} + 4x + 3} \right)\left[ { - 2{{\left( {x + 1} \right)}^{ - 3}}} \right] + {\left( {x + 1} \right)^{ - 2}}\left( {4x + 4} \right) \cr
& f\left( x \right) = - 2{\left( {x + 1} \right)^{ - 3}}\left( {2{x^2} + 4x + 3} \right) + 4{\left( {x + 1} \right)^{ - 1}} \cr
& {\text{Simplifying}} \cr
& f\left( x \right) = - 2{\left( {x + 1} \right)^{ - 3}}\left[ {2{x^2} + 4x + 3 - 2{{\left( {x + 1} \right)}^2}} \right] \cr
& f\left( x \right) = - 2{\left( {x + 1} \right)^{ - 3}}\left( {2{x^2} + 4x + 3 - 2{x^2} - 4x - 2} \right) \cr
& f\left( x \right) = - 2{\left( {x + 1} \right)^{ - 3}}\left( 1 \right) \cr
& f\left( x \right) = - \frac{2}{{{{\left( {x + 1} \right)}^3}}} \cr} $$
