The two curves x^3-3 x y^2+2=0 and 3 x^2 y-y^3=2

We have,

$x^3-3 x y^2+2=0$              .......(i)

and, $3 x^2 y-y^3-2=0$        ......(ii)

Differentiating (i) and (ii) with respect to x, we obtain

$\left(\frac{d y}{d x}\right)_{C_1}=\frac{x^2-y^2}{2 x y}$  and  $\left(\frac{d y}{d x}\right)_{C_2}=\frac{-2 x y}{x^2-y^2}$

Clearly, $\left(\frac{d y}{d x}\right)_{C_1} \times\left(\frac{d y}{d x}\right)_{C_2}=-1$

Hence, the two curves cut at right angles.