Answer
$$\,\,\,\,\,y = - \frac{4}{{\sqrt 6 }}x + \frac{2}{{\sqrt 3 }} + \frac{\pi }{3}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\cos ^{ - 1}}{x^2};{\text{ }}\left( {\frac{1}{{\sqrt 2 }},\frac{\pi }{3}} \right) \cr
& {\text{differentiate }}f\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}{x^2}} \right] \cr
& {\text{use }}\frac{d}{{du}}\left[ {{{\cos }^{ - 1}}u} \right] = - \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}} \cr
& f'\left( x \right) = - \frac{1}{{\sqrt {1 - {{\left( {{x^2}} \right)}^2}} }}\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& f'\left( x \right) = - \frac{1}{{\sqrt {1 - {x^4}} }}\left( {2x} \right) \cr
& f'\left( x \right) = - \frac{{2x}}{{\sqrt {1 - {x^4}} }} \cr
& \cr
& {\text{find the slope at the point }}\left( {\frac{1}{{\sqrt 2 }},\frac{\pi }{3}} \right) \cr
& m = f'\left( {\frac{1}{{\sqrt 2 }}} \right) = - \frac{{2\left( {1/\sqrt 2 } \right)}}{{\sqrt {1 - {{\left( {1/\sqrt 2 } \right)}^4}} }} \cr
& \,\,m = - \frac{2}{{\sqrt 2 \sqrt {3/4} }} \cr
& \,\,m = - \frac{4}{{\sqrt 6 }} \cr
& \,{\text{find the equation of the tangent line at the point }}\left( {\frac{1}{{\sqrt 2 }},\frac{\pi }{3}} \right) \cr
& \,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,\,\,\,y - \frac{\pi }{3} = - \frac{4}{{\sqrt 6 }}\left( {x - \frac{1}{{\sqrt 2 }}} \right) \cr
& \,\,\,\,\,y - \frac{\pi }{3} = - \frac{4}{{\sqrt 6 }}x + \frac{4}{{2\sqrt 3 }} \cr
& \,\,\,\,\,y = - \frac{4}{{\sqrt 6 }}x + \frac{2}{{\sqrt 3 }} + \frac{\pi }{3} \cr} $$
