The two curves $x^3-3 x y^2+15=0$ and $3 x^2 y-y^3+17=0$ :

cut at right angles

Curves

$x^3-3 x y^2+15=0$        ...(1)

$3 x^2 y-y^3+17=0$         ...(2)

Differentiating (1) wrt (x)

$\frac{d}{d x}\left(x^3-3 y^2+15\right)=0$

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

So  $6 x y \frac{d y}{d x}=-3\left(y^2-x^2\right)$

$\frac{d y}{d x}=\frac{-3\left(y^2-x^2\right)}{6 x y}=m_1$

Slope of curve 1

Differentiating (2) wrt (x)

$\frac{d}{d x}\left(3 x^2 y-y^3+17\right)=0$

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

$6 x y=\frac{d y}{d x}\left(3 y^2-3 x^2\right)$

$\frac{d y}{d x}=\frac{6 x y}{3\left(y^2-x^2\right)}=m_2$

slope of curve 2

$m_1 m_2=\frac{-3\left(y^2-x^2\right)}{6 x y} \frac{6 x y}{3\left(y^2-x^2\right)}=-1$

m₁m₂ = -1