Answer
$$f'\left( x \right) = \frac{{{{\sec }^2}x - {{\tan }^2}x}}{{{{\left( {1 + x\tan x} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{\sin x\sec x}}{{1 + x\tan x}} \cr
& {\text{identity }}\sec x = \frac{1}{{\cos x}} \cr
& f\left( x \right) = \frac{{\sin x\left( {\frac{1}{{\cos x}}} \right)}}{{1 + x\tan x}} \cr
& f\left( x \right) = \frac{{\frac{{\sin x}}{{\cos x}}}}{{1 + x\tan x}} \cr
& {\text{ identity }}\frac{{\sin x}}{{\cos x}} = \tan x \cr
& f\left( x \right) = \frac{{\tan x}}{{1 + x\tan x}} \cr
& {\text{quotient rule }} \cr
& \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr
& f'\left( x \right) = \frac{{\left( {1 + x\tan x} \right)\left( {\tan x} \right)' - \left( {\tan x} \right)\left( {1 + x\tan x} \right)'}}{{{{\left( {1 + x\tan x} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{\left( {1 + x\tan x} \right)\left( {{{\sec }^2}x} \right) - \left( {\tan x} \right)\left( {x{{\sec }^2}x + \tan x} \right)}}{{{{\left( {1 + x\tan x} \right)}^2}}} \cr
& {\text{simplify}} \cr
& f'\left( x \right) = \frac{{{{\sec }^2}x + x\tan x{{\sec }^2}x - x\tan x{{\sec }^2}x - {{\tan }^2}x}}{{{{\left( {1 + x\tan x} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{{{\sec }^2}x - {{\tan }^2}x}}{{{{\left( {1 + x\tan x} \right)}^2}}} \cr} $$
