The point of intersection of the curves $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and $x^{2}+y^{2}=3$ in the first quadrant is $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$

Now slope of tangent to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ at $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$ is

$\mathrm{m}_{1}=-\frac{1}{3 \sqrt{3}}$

And slope of tangent to the circle at $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$ is $\mathrm{m}_{2}$

$=-\sqrt{3}$

So, if angle between both curves is $\theta$ then

$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|=\left|\frac{-\frac{1}{3 \sqrt{3}}+\sqrt{3}}{1+\left(-\frac{1}{3 \sqrt{3}}(-\sqrt{3})\right)}\right|$$=\frac{2}{\sqrt{3}}$