Find $\frac{d y}{d x}$ :

$x^{3}+x^{2} y+x y^{2}+y^{3}=81$


The given relationship is $x^{3}+x^{2} y+x y^{2}+y^{3}=81$

Differentiating this relationship with respect to x, we obtain

$\frac{d}{d x}\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)=\frac{d}{d x}(81)$

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

$\Rightarrow 3 x^{2}+\left[y \frac{d}{d x}\left(x^{2}\right)+x^{2} \frac{d y}{d x}\right]+\left[y^{2} \frac{d}{d x}(x)+x \frac{d}{d x}\left(y^{2}\right)\right]+3 y^{2} \frac{d y}{d x}=0$

$\Rightarrow 3 x^{2}+\left[y \cdot 2 x+x^{2} \frac{d y}{d x}\right]+\left[y^{2} \cdot 1+x \cdot 2 y \cdot \frac{d y}{d x}\right]+3 y^{2} \frac{d y}{d x}=0$

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

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

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