Given curves are


On solving Eqs. (i) and (ii), we get
$$
\therefore \quad x= \pm \sqrt{3}, y=1
$$
Thus, their points of intersection are $(\sqrt{3}, 1)$ and $(-\sqrt{3}, 1)$. Now, from Eq. (i), $\frac{d y}{d x}=\frac{2 x}{3}$ and from Eq. (ii), $\frac{d y}{d x}=-\frac{x}{y}$
To calculate angle, any of the point can be taken so we are taking $(\sqrt{3}, 1)$ as the point of intersection. Now, let $m_1$ and $m_2$ be the slope of tangent to the curves at $(\sqrt{3}, 1)$. Then,
$$
m_1=\frac{2 \sqrt{3}}{3} \text { and } m_2=-\sqrt{3}
$$
Now, the angle between two curves
$\begin{aligned} & \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{\frac{2 \sqrt{3}}{3}-(-\sqrt{3})}{1+\frac{2 \sqrt{3}}{3} \times(-\sqrt{3})}\right| \\ & =\left|\frac{\left(\frac{5 \sqrt{3}}{3}\right)}{(-1)}\right|=\frac{5}{\sqrt{3}} \\ & \Rightarrow \quad \theta=\tan ^{-1}\left(\frac{5}{\sqrt{3}}\right)\end{aligned}$