Answer
$$x = - 3,{\text{ }}x = - 3 + \sqrt 6 ,{\text{ }}x = - 3 - \sqrt 6 $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{{\left( {x - 3} \right)}^4}}}{{{x^2} + 2x}} \cr
& {\text{Find }}f'\left( x \right){\text{ using the product rule}} \cr
& f'\left( x \right) = \frac{{\left( {{x^2} + 2x} \right)\left( 4 \right){{\left( {x - 3} \right)}^3} - {{\left( {x - 3} \right)}^4}\left( {2x + 2} \right)}}{{{{\left( {{x^2} + 2x} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{4x\left( {x + 2} \right){{\left( {x - 3} \right)}^3} - 2\left( {x + 1} \right){{\left( {x - 3} \right)}^4}}}{{{{\left( {{x^2} + 2x} \right)}^2}}} \cr
& {\text{Factoring the numerator}} \cr
& f'\left( x \right) = \frac{{2{{\left( {x - 3} \right)}^3}\left[ {2x\left( {x + 2} \right) - \left( {x + 1} \right){{\left( {x - 3} \right)}^4}} \right]}}{{{{\left( {{x^2} + 2x} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{2{{\left( {x - 3} \right)}^3}\left( {2{x^2} + 4x - {x^2} + 2x + 3} \right)}}{{{{\left( {{x^2} + 2x} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{2{{\left( {x - 3} \right)}^3}\left( {{x^2} + 6x + 3} \right)}}{{{{\left( {{x^2} + 2x} \right)}^2}}} \cr
& {\text{Set the derivative }}f'\left( x \right) = 0 \cr
& \frac{{2{{\left( {x - 3} \right)}^3}\left( {{x^2} + 6x + 3} \right)}}{{{{\left( {{x^2} + 2x} \right)}^2}}} = 0 \cr
& 2{\left( {x - 3} \right)^3}\left( {{x^2} + 6x + 3} \right) = 0 \cr
& x - 3 = 0,{\text{ }}{x^2} + 6x + 3 = 0 \cr
& x = 3,{\text{ }}{x^2} + 6x + 3 = 0 \cr
& {\text{Solving }}{x^2} + 6x + 3 = 0{\text{ by the quadratic formula}} \cr
& x = \frac{{ - 6 \pm \sqrt {{{\left( 6 \right)}^2} - 4\left( 1 \right)\left( 3 \right)} }}{2} \cr
& x = \frac{{ - 6 \pm \sqrt {24} }}{2} \cr
& x = \frac{{ - 6 \pm 2\sqrt 6 }}{2} \cr
& x = - 3 \pm \sqrt 6 \cr
& {\text{At these points the curve has a horizontal tangent line}}{\text{.}} \cr
& {\text{Therefore, the points are:}} \cr
& x = - 3,{\text{ }}x = - 3 + \sqrt 6 ,{\text{ }}x = - 3 - \sqrt 6 \cr} $$
