Answer
$$\left( {{f^{ - 1}}} \right)'\left( x \right) = - \frac{2}{{{x^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{2}{{x + 3}} \cr
& f'\left( x \right) = - \frac{2}{{{{\left( {x + 3} \right)}^2}}} \cr
& {\text{Calculating the inverse of the function}} \cr
& y = \frac{2}{{x + 3}} \cr
& x = \frac{2}{{y + 3}} \cr
& x\left( {y + 3} \right) = 2 \cr
& xy + 3x = 2 \cr
& xy = 2 - 3x \cr
& y = \frac{2}{x} - 3 \cr
& {f^{ - 1}}\left( x \right) = \frac{2}{x} - 3 \cr
& {\text{By the formula 2: }}\left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{1}{{f'\left( {{f^{ - 1}}\left( x \right)} \right)}} \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{1}{{ - \frac{2}{{{{\left( {\frac{2}{x} - 3 + 3} \right)}^2}}}}} \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{1}{{ - \frac{2}{{{{\left( {\frac{2}{x}} \right)}^2}}}}} \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = - \frac{{{{\left( {\frac{2}{x}} \right)}^2}}}{2} = - \frac{1}{{\frac{2}{{4{x^2}}}}} \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = - \frac{2}{{{x^2}}} \cr
& \cr
& {\text{Differentiating }}\left( {{f^{ - 1}}} \right)\left( x \right){\text{ directly}} \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{2}{x} - 3} \right] \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = - \frac{2}{{{x^2}}} \cr} $$
