Answer
$-\displaystyle \frac{1}{2x^{2}}\cos^{-1}(\frac{1}{x^{2}})+\frac{1}{2}\sqrt{1-\frac{1}{x^{4}}}+C$
Work Step by Step
$\displaystyle \int\frac{\cos^{-1}(x^{-2})}{x^{3}}dx=\int\cos^{-1}(x^{-2})x^{-3}dx=\quad \left[\begin{array}{l}
u=x^{-2}\\
du=-2x^{-3}dx
\end{array}\right]$
$=-\displaystyle \frac{1}{2}\int\cos^{-1}udu$
Table of integrals:
$\color{blue}{ 88. \quad \displaystyle \int\cos^{-1}udu=u\cos^{-1}u-\sqrt{1-u^{2}}+C }$
$=-\displaystyle \frac{1}{2}(u\cos^{-1}u-\sqrt{1-u^{2}})+C$
... bring back x...
$=-\displaystyle \frac{1}{2x^{2}}\cos^{-1}(\frac{1}{x^{2}})+\frac{1}{2}\sqrt{1-\frac{1}{x^{4}}}+C$