Answer
$$y' = {e^{x\tan x}}\left( {x{{\sec }^2}x + \tan x} \right)$$
Work Step by Step
$$\eqalign{
& y = {e^{x\tan x}} \cr
& {\text{differentiate}} \cr
& y' = \left( {{e^{x\tan x}}} \right)' \cr
& {\text{use the chain rule}} \cr
& y' = {e^{x\tan x}}\left( {x\tan x} \right)' \cr
& {\text{product rule}} \cr
& y' = {e^{x\tan x}}\left( {x\left( {\tan x} \right)' + \tan x\left( x \right)'} \right) \cr
& y' = {e^{x\tan x}}\left( {x{{\sec }^2}x + \tan x\left( 1 \right)} \right) \cr
& {\text{Simplify}} \cr
& y' = {e^{x\tan x}}\left( {x{{\sec }^2}x + \tan x} \right) \cr} $$