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

We are given that, ${x^3} + {x^2}y + x{y^2} + {y^2} = 81$ …(i)
Differentiating (i) on both sides w.r.t. x, we get

$3{x^2} + {x^2}\cfrac{{dy}}{{dx}} + y(2x) + {y^2} + x\left( {2y\cfrac{{dy}}{{dx}}} \right) + 3{y^2}\cfrac{{dy}}{{dx}} = 0$

$\Rightarrow$ $\cfrac{{dy}}{{dx}}[{x^2} + 2xy + 3{y^2}] = - (3{x^2} + 2xy + {y^2})$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{{ - (3{x^2} + 2xy + {y^2})}}{{{x^2} + 2xy + 3{y^2}}}$