(a) $\left(e^{\frac{1}{3}}, \frac{1}{3 e}\right) $

(b) $ \frac{2}{9} -\frac{\ln 3}{18}$

(a)
$$
\begin{array}{l}
y=\frac{\ln x}{x^3} \\
\frac{d y}{d x}=\frac{\frac{1}{x} \cdot x^3-\ln x \cdot 3 x^2}{x^6} \\
\frac{d y}{d x}=\frac{x^2(1-3 \ln x)}{x^6} \\
\frac{d y}{d x}=\frac{1-3 \ln x}{x^4}
\end{array}
$$
Let $\frac{d y}{d x}=0$, then $1-3 \ln x=0 \Rightarrow x=e^{\frac{1}{3}}$ and $y=\frac{1}{3 e}$
Coordinates are $\left(e^{\frac{1}{3}}, \frac{1}{3 e}\right)$.
(b)
$$
\begin{array}{l}
\int_1^3 x^{-3} \ln x d x \\
=\left[-\frac{1}{2} x^{-2} \cdot \ln x\right]_1^3-\int_1^3-\frac{1}{2} x^{-2} \cdot \frac{1}{x} d x \\
=-\frac{\ln 3}{18}+\frac{1}{2} \int_1^3 x^{-3} d x \\
=-\frac{\ln 3}{18}+\frac{1}{2}\left[-\frac{1}{2} x^{-2}\right]_1^3 \\
=-\frac{\ln 3}{18}+\frac{1}{2}\left(-\frac{1}{18}+\frac{1}{2}\right) \\
= \frac{2}{9} -\frac{\ln 3}{18}
\end{array}
$$