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