Answer
$$\frac{d^2y}{dx^2}=\frac{2}{x^3}\sec^2\frac{1}{x}\tan\frac{1}{x}$$
Work Step by Step
$$\frac{dy}{dx}=\Big(x\tan\frac{1}{x}\Big)'=x'\tan\frac{1}{x}+x\Big(\tan\frac{1}{x}\Big)'=
\tan\frac{1}{x}+x\sec^2\frac{1}{x}\Big(\frac{1}{x}\Big)'=\tan\frac{1}{x}+x\sec^2\frac{1}{x}\Big(-\frac{1}{x^2}\Big)=
\tan\frac{1}{x}-\frac{1}{x}\sec^2\frac{1}{x}$$
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(\frac{dy}{dx}\Big)=\Big(\tan\frac{1}{x}-\frac{1}{x}\sec^2\frac{1}{x}\Big)'=
\sec^2\frac{1}{x}\Big(\frac{1}{x}\Big)'-\Big(\Big(\frac{1}{x}\Big)'\sec^2\frac{1}{x}+\frac{1}{x}\Big(\sec^2\frac{1}{x}\Big)'\Big)=
\sec^2\frac{1}{x}\Big(-\frac{1}{x^2}\Big)-\Big(-\frac{1}{x^2}\sec^2\frac{1}{x}\Big)-\frac{1}{x}\cdot2\sec\frac{1}{x}\Big(\sec\frac{1}{x}\Big)'=
-\frac{2}{x}\sec\frac{1}{x}\sec\frac{1}{x}\tan\frac{1}{x}\Big(\frac{1}{x}\Big)'=
-\frac{2}{x}\sec^2\frac{1}{x}\tan\frac{1}{x}\Big(-\frac{1}{x^2}\Big)=
\frac{2}{x^3}\sec^2\frac{1}{x}\tan\frac{1}{x}$$
