$f(x)=\frac{x}{8}+\frac{2}{x}$
$\therefore \quad f^{\prime}(x)=\frac{1}{8}-\frac{2}{x^{2}}=\frac{x^{2}-16}{8 x^{2}}$
For maximum or minimum $f^{\prime}(x)$ must be vanish.
$$
\begin{array}{ll}
\therefore & f^{\prime}(x)=0 \\
\Rightarrow & \frac{x^{2}-16}{8 x^{2}}=0 \Rightarrow x=4,-4 \\
& x \in[1,6] \\
\therefore & x \neq-4
\end{array}
$$
Also, in $[1,4], f^{\prime}(x) < 0 \Rightarrow f(x)$ is decreasing. $\ln [4,6], f^{\prime}(x)>0 \Rightarrow f(x)$ is increasing.
$f(1)=\frac{1}{8}+\frac{2}{1}=\frac{17}{8}$
$f(8)=\frac{6}{8}+\frac{2}{6}=\frac{3}{4}+\frac{1}{3}=\frac{13}{12}$
Hence, maximum value of $f(x)$ in [1,6] is $\frac{17}{8}$.