Point of intersection of the curves
$$
\begin{aligned}
\Rightarrow \quad x^2+4 y & =0 \text { and } x y=2 \\
\Rightarrow \quad x^2 & =-4 y \text { and } y=\frac{2}{x} \\
\Rightarrow \quad x^2 & =\frac{-8}{x} \\
x^3 & =-8 \Rightarrow x=-2
\end{aligned}
$$

Slope of first curve at $x=-2$
$$
\begin{aligned}
2 x+4 \frac{d y}{d x} & =0 \Rightarrow \frac{d y}{d x}=-\frac{x}{2} \\
m_1 & =\left(\frac{d y}{d x}\right)_{x=-2}=\frac{-(-2)}{2}=1
\end{aligned}
$$

Slope of the second curve $x y=2$
$$
\begin{aligned}
y & =\frac{2}{x} \\
\frac{d y}{d x} & =\frac{-2}{x^2} \\
m_2 & =\left(\frac{d y}{d x}\right)_{x=-2}=\frac{-2}{(-2)^2}=\frac{-1}{2}
\end{aligned}
$$

If angle between them $\theta$, then
$$
\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{1-\left(-\frac{1}{2}\right)}{1+(1)\left(-\frac{1}{2}\right)}\right|=\left|\frac{3 / 2}{1 / 2}\right|=3
$$