The two curves \( x^{3}-3 x y^{2}+2=0 \) and \( 3 x^{2} y-y^{3}=2 \)

Given curves, \(\rightarrow\)
\(x^{3}-3 x y^{2}+2=0 \rightarrow(1)\)
\(3 x^{2} y-y^{3}=2 \rightarrow(2)\)
Differentiating Eqs. (1) and (2) both the sides with respect to \(x\), we get
\(3 x^{2}-3\left(y^{2}+x\left(2 y y^{\prime}\right)\right)=0\)
\(\Rightarrow x^{2}=y^{2}+2 x y y^{\prime}\)
\(\Rightarrow y^{\prime}=\frac{x^{2}-y^{2}}{2 x y}=m_{1}(\) Let \() \rightarrow(3)\)
\(3\left(x^{2} y^{\prime}+2 x y\right) 3 y^{2} y=0\)
\(\Rightarrow x^{2} y^{\prime}+2 x y-y^{2} y^{\prime}=0\)
\(\Rightarrow y^{\prime}=\frac{-2 x y}{x^{2}-y^{2}}=m_{2}(\) Let \() \rightarrow(4)\)
From Eqs. (3) and (4), we get
\(m_{1} m_{2}=-1\)
Hence, the two curves cut at an angle \(\frac{\Pi}{2}\) i.e. right angle