The two curves x^3-3 x y^2+5=0 and 3 x^2 y-y^3-7=0

Differentiating $x^3-3 x y^2+5=0$, we get

$3 x^2-3 y^2-6 x y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{x^2-y^2}{2 x y}$

Differentiating $3 x^2 y-y^3-7=0$, we get

$6 x y+3 x^2 \frac{d y}{d x}-3 y^2 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{2 x y}{y^2-x^2}$

Since, product of slopes is $\frac{x^2-y^2}{2 x y} \times \frac{2 x y}{y^2-x^2}=-1$

∴ The two curves cut at right angle.