$\begin{aligned} 8 f(x)+6 f\left(\frac{1}{x}\right) & =x+5 \\ y & =x^2 f(x)\end{aligned}$
Replace $x$ by $\frac{1}{x}$ in Eq. (i), we obtain
$$
8 f\left(\frac{1}{x}\right)+6 f(x)=\frac{1}{x}+5
$$

Solve Eqs. (i) and (ii) for $f(x)$ and $f\left(\frac{1}{x}\right)$
Multiply by 8 in Eq. (i) and by 6 in Eq. (ii) and subtract, we get
$$
\begin{aligned}
& 64 f(x)-36 f(x)=8 x-\frac{6}{x}+10 \\
& \Rightarrow \quad 28 f(x)=8 x-\frac{6}{x}+10
\end{aligned}
$$

Now, $28 x^2 f(x)=8 x^3-6 x+10 x^2$
$$
\Rightarrow \quad 28 y=8 x^3-6 x+10 x^2
$$

Differentiating it,
$$
\Rightarrow \begin{aligned}
28 \frac{d y}{d x} & =24 x^2-6+20 x \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{24 x^2-6+20 x}{28} \\
\left(\frac{d y}{d x}\right)_{x=-1} & =\frac{24(-1)^2-6+20(-1)}{28}=-\frac{1}{14}
\end{aligned}
$$