Answer
$$k'(x)=\sec\frac{1}{x}\Big(2x-\tan\frac{1}{x}\Big)$$
Work Step by Step
$$k(x)=x^2\sec\Big(\frac{1}{x}\Big)$$
The derivative of function $k(x)$ is: $$k'(x)=(x^2)'\sec\Big(\frac{1}{x}\Big)+x^2\Bigg[\sec\Big(\frac{1}{x}\Big)\Bigg]'$$
$$k'(x)=2x\sec\frac{1}{x}+x^2\sec\frac{1}{x}\tan\frac{1}{x}\Big(\frac{1}{x}\Big)'$$
$$k'(x)=2x\sec\frac{1}{x}+x^2\sec\frac{1}{x}\tan\frac{1}{x}\Big(\frac{-1(x)'}{x^2}\Big)$$
$$k'(x)=2x\sec\frac{1}{x}+x^2\sec\frac{1}{x}\tan\frac{1}{x}\Big(-\frac{1}{x^2}\Big)$$
$$k'(x)=2x\sec\frac{1}{x}-\sec\frac{1}{x}\tan\frac{1}{x}$$
$$k'(x)=\sec\frac{1}{x}\Big(2x-\tan\frac{1}{x}\Big)$$