Answer
$$y' = \frac{1}{{\left( {{x^2} - 1} \right){{\left( {{{\tanh }^{ - 1}}x} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{{{\tanh }^{ - 1}}x}} \cr
& {\text{write with positive exponent}}{\text{, recall }}\frac{1}{{{z^n}}} = {z^{ - n}} \cr
& y = {\left( {{{\tanh }^{ - 1}}x} \right)^{ - 1}} \cr
& {\text{find the derivatvive by using the chain rule}} \cr
& y' = - {\left( {{{\tanh }^{ - 1}}x} \right)^{ - 2}}\left( {{{\tanh }^{ - 1}}x} \right)' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.5 }}\left( {{\text{ see page 479}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 - {u^2}}}\frac{{du}}{{dx}} \cr
& y' = - {\left( {{{\tanh }^{ - 1}}x} \right)^{ - 2}}\left( {\frac{1}{{1 - {x^2}}}} \right) \cr
& {\text{simplifying}} \cr
& y' = {\left( {{{\tanh }^{ - 1}}x} \right)^{ - 2}}\left( {\frac{1}{{{x^2} - 1}}} \right) \cr
& y' = \frac{1}{{\left( {{x^2} - 1} \right){{\left( {{{\tanh }^{ - 1}}x} \right)}^2}}} \cr} $$
