Find $\frac{d y}{d x}$ of the following implicit functions : $y^{3} - 3 y^{2} x = x^{3} + 3 x^{2} y$

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Solution

$y^{3} - 3 y^{2} x = x^{3} + 3 x^{2} y$
Differentiating both sides w.r.t $x$
$3 y^{2} \frac{d y}{d x} - 3 \times 2 y \frac{d y}{d x} . x - 3 y^{2} = 3 x^{2} \frac{d y}{d x} + 3 x^{2} 3 y^{2} \frac{d y}{d x} - 3 x^{2} \frac{d y}{d x} - 6 x y \frac{d y}{d x} = 3 x^{2} + 3 y^{2} \frac{d y}{d x} ({3 y}^{2} - 3 x^{2} - 6 x y) = 3 x^{2} + {3 y}^{2} \frac{d y}{{d x}^{2}} = \frac{3 x^{2} + 3 y^{2}}{{3 y}^{2} - {3 x}^{2} - 6 x y} = \frac{x^{2} + y^{2}}{y^{2} - x^{2} - 2 x y}$

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