Question

Find $\dfrac{{dy}}{{dx}}{\text{ of }}{x^3} + {x^2}y + x{y^2} + {y^3} = 81$

Given equation, ${x^3} + {x^2}y + x{y^2} + {y^3} = 81$
Differentiate both sides with respect to $x$, we get
$\dfrac{{d\left( {{x^3} + {x^2}y + x{y^2} + {y^3}} \right)}}{{dx}} = \dfrac{{d\left( {81} \right)}}{{dx}} \\ \Rightarrow 3{x^2} + \left( {y.2x + {x^2}\dfrac{{dy}}{{dx}}} \right) + \left( {{y^2}.1 + x.2y.\dfrac{{dy}}{{dx}}} \right) + 3{y^2}\dfrac{{dy}}{{dx}} = 0 \\$
$\Rightarrow \left( {{x^2} + 2xy + 3{y^2}} \right)\dfrac{{dy}}{{dx}} + \left( {3{x^2} + 2xy + {y^2}} \right) = 0 \\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \left( {3{x^2} + 2xy + {y^2}} \right)}}{{\left( {{x^2} + 2xy + 3{y^2}} \right)}} \\$