Answer
$$f'\left( x \right) = 4\tan x + 2\ln \left( {\sec x\csc x} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \left( {{{\sec }^4}x{{\tan }^2}x} \right) \cr
& {\text{Using the logarithmic properties}} \cr
& f\left( x \right) = \ln \left( {{{\sec }^4}x} \right) + \ln \left( {{{\tan }^2}x} \right) \cr
& f\left( x \right) = 4\ln \left( {\sec x} \right) + 2\ln \left( {\tan x} \right) \cr
& {\text{Differentiate both sides with respect to }}x \cr
& f'\left( x \right) = 4\left( {\frac{{\sec x\tan x}}{{\sec x}}} \right) + 2\ln \left( {\frac{{{{\sec }^2}x}}{{\tan x}}} \right) \cr
& f'\left( x \right) = 4\tan x + 2\ln \left( {\frac{1}{{{{\cos }^2}x}}\frac{{\cos x}}{{\sin x}}} \right) \cr
& f'\left( x \right) = 4\tan x + 2\ln \left( {\frac{1}{{\cos x}}\frac{1}{{\sin x}}} \right) \cr
& f'\left( x \right) = 4\tan x + 2\ln \left( {\sec x\csc x} \right) \cr} $$
