Solution:- \(x^{3}+y^{3}-1=0\) \(x^{3}+y^{3}=1\) now, \(\frac{d}{dx}(x^{3}+y^{3})=\frac{d}{dx}(1)\) \(\frac{d}{dx}(x^{3})\)+ \(\frac{d}{dx}(y^{3})= 0\) \(3x^{2}+3y^{2} \frac{d}{dx}=0\) \(3y^{2}\frac{d}{dx}= – 3x^{2}\) \(\frac{d}{dx}\)\(=\)\(-\)\(\frac{x^{2}}{y^{2}}\)