Angle

$C_1:x^3-3xy^2+2=0$ $C_2:3x^2y-y^3-2=0$ Solving $C_1$ and $C_2$ by adding them:

$(x-y)^3=0\implies x=y$

Putting $x=y$ in curves we get point of contact is $(1,1)$ .

Now, for $C_1$ :-

$\left(\dfrac{dy}{dx}\right)_{(1,1)}=\left(\dfrac{x^2-y^2}{2xy}\right)_{(1,1)}=\tan 0$

For $C_2$ :-

$\left(\dfrac{dy}{dx}\right)_{(1,1)}=\left(\dfrac{2xy}{y^2-x^2}\right)_{(1,1)}=\tan\left(\dfrac{\pi}2\right)$

And we know angle b/w curves is the angle b/w their respective tangents at point of contact which here is:- $\Large \dfrac{\pi}2-0=\boxed{\color{#007fff}{\dfrac{\pi}2}}$