Answer
$\sec\dfrac{1}{x}(2x-\tan\dfrac{1}{x})$
Work Step by Step
Consider the derivative of function $k(x)$ as follow:
$k'(x)=(x^2)'\sec(\dfrac{1}{x})+x^2[\sec(\dfrac{1}{x})]')=2x\sec\dfrac{1}{x}+x^2\sec\dfrac{1}{x}\tan\dfrac{1}{x}(\dfrac{1}{x})$
This implies that
$k'(x)=2x\sec(\dfrac{1}{x})+x^2\sec(\dfrac{1}{x})\tan(\dfrac{1}{x})(-\dfrac{1}{x^2})=\sec\dfrac{1}{x}(2x-\tan\dfrac{1}{x})$
