We have,
$$
x y=1
$$
and $x^2+8 y=0$
On solving Eqs. (i) and (ii), we get
$$
x=-2, y=-\frac{1}{2}
$$
Now, $x y=1$
$$
\Rightarrow \frac{d y}{d x}=-\frac{y}{x} \Rightarrow\left(\frac{d y}{d x}\right)_{\left(-2, \frac{-1}{2}\right)}=-\frac{1}{4}=m_1
$$
and $x^2+8 y=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{4}$
$$
\begin{aligned}
& \Rightarrow \quad\left(\frac{d y}{d x}\right)_{\left(-2,-\frac{1}{2}\right)}=\frac{1}{2}=m_2 \\
& \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \Rightarrow \tan \theta=\left|\frac{-\frac{1}{4}-\frac{1}{2}}{1-\frac{1}{8}}\right| \\
& \therefore \tan \theta=\frac{6}{7}
\end{aligned}
$$