Answer
$$f'\left( x \right) = - \frac{{{e^x}{{\sec }^2}{e^x}}}{{\left| {\tan {e^x}} \right|\sqrt {{{\tan }^2}{e^x} - 1} }}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\csc ^{ - 1}}\left( {\tan {e^x}} \right) \cr
& {\text{find the derivative}}{\text{,}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\csc }^{ - 1}}\left( {\tan {e^x}} \right)} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{{\csc }^{ - 1}}u} \right] = - \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}}.{\text{ let }}u = \tan {e^x} \cr
& f'\left( x \right) = - \frac{1}{{\left| {\tan {e^x}} \right|\sqrt {{{\left( {\tan {e^x}} \right)}^2} - 1} }}\frac{d}{{dx}}\left[ {\tan {e^x}} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {\tan u} \right] = {\sec ^2}u\frac{{du}}{{dx}} \cr
& f'\left( x \right) = - \frac{1}{{\left| {\tan {e^x}} \right|\sqrt {{{\left( {\tan {e^x}} \right)}^2} - 1} }}\left[ {{{\sec }^2}{e^x}} \right]\frac{d}{{dx}}\left[ {{e^x}} \right] \cr
& {\text{simplifying}} \cr
& f'\left( x \right) = - \frac{1}{{\left| {\tan {e^x}} \right|\sqrt {{{\tan }^2}{e^x} - 1} }}\left( {{{\sec }^2}{e^x}} \right)\left( {{e^x}} \right) \cr
& f'\left( x \right) = - \frac{{{e^x}{{\sec }^2}{e^x}}}{{\left| {\tan {e^x}} \right|\sqrt {{{\tan }^2}{e^x} - 1} }} \cr} $$
