The general term in $\left(a x^{2}+\frac{1}{b x}\right)^{13}$ is
$T_{r+1}={ }^{13} C_{r}\left(a x^{2}\right)^{13-r}\left(\frac{1}{b x}\right)^{r}$
$$
={ }^{13} C_{r} a^{13-r} \times b^{-r}(x)^{26-3 r}
$$
For coefficient of $x^{8}$. put $26-3 r=8$
$\Rightarrow \quad 3 r=18 \Rightarrow r=6$
$\therefore$
$T_{7}={ }^{13} C_{6} a^{13-6} b^{-6} x^{8}$
$={ }^{13} C_{6} a^{7} b^{-6} x^{8}$
Now, the general term in $\left(a x-\frac{1}{b x^{2}}\right)^{13}$ is
$$
\begin{aligned}
T_{r+1}^{'} &={ }^{13} C_{r}(a x)^{13-r}\left(-\frac{1}{b x^{2}}\right)^{r} \\
&={ }^{13} C_{r}^{*} a^{13-r} \times b^{-r} \times(x)^{13-3 r}(-1)^{\prime}
\end{aligned}
$$
For coefficient of $x^{-8}$, put $13-3 r=-8$
According to the given condition, Coefficient of $x^{8}$ in $\left(a x^{2}+\frac{1}{b x}\right)^{13}$
$$
\begin{array}{l}
=\text { Coefficient of } x^{-8} \text {in }\left(a x-\frac{1}{b x^{2}}\right)^{13} \\
\therefore{ }^{13} C_{6} a^{7} b^{-6}=-{ }^{13} C_{7} a^{5} b^{-7} \\
\Rightarrow{ }^{13} C_{7} \frac{a^{7}}{b^{6}}=-{ }^{13} C_{7} \frac{a^{6}}{b^{7}} \\
\Rightarrow \quad \frac{a^{7}}{a^{6}}=-\frac{b^{6}}{b^{7}} \Rightarrow \quad a=-\frac{1}{b} \Rightarrow a b+1=0
\end{array}
$$