Answer
$$ \frac{-1}{x}\sin \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) +c$$
Work Step by Step
Given
$$\int\frac{\cos(1/x)}{x^3}dx$$
let $z=\dfrac{1}{x}\ \ \to \ dz=\dfrac{-1}{x^2}dx$,
then
$$\int\frac{\cos(1/x)}{x^3}dx = -\int z\cos zdz $$
Let
\begin{align*}
u&=z\ \ \ \ \ \ \ \ \ \ \ \ dv=\cos z dz\\
du&=dz\ \ \ \ \ \ \ \ \ \ \ \ v=\sin z
\end{align*}
Then
\begin{align*}
\int\frac{\cos(1/x)}{x^3}dx &=- \int z\cos zdz \\
&=-z\sin z +\int \sin z dz\\
&=-z\sin z - \cos z +c\\
&= \frac{-1}{x}\sin \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) +c
\end{align*}