# Find $$\frac {dy}{dx}$$ in the following: $$x^3 +x^2y + xy^2 +y^3 = 81$$

$\begin{array}{1 1} -\large \frac{y^2+2xy+3x^3}{x^2+2xy+3y^2} \\\large \frac{y^2+2xy+3x^3}{x^2+2xy+3y^2} \\ -\large \frac{x^2+2xy+3y^2}{y^2+2xy+3x^3} \\ \large \frac{x^2+2xy+3y^2}{y^2+2xy+3x^3} \end{array}$

Toolbox:
• For equations that are not of the form $y = f(x)$, we need to differentiate each term separately on LHS and RHS and then calculate $\large \frac{dy}{dx}$
Given $x^3+x^2y+xy^2+y^3=81$
This is not a standard differentiation of the form $y = f(x)$. We need to differentiate each term separately on LHS and RHS and then calculate $\large \frac{dy}{dx}$
According to the Product Rule for differentiation, given two functions $u$ and $v, \large \frac {d(uv)}{dx} $$= u \large \frac{dv}{dx}$$+ v \large \frac{du}{dx}$
Differentiating both sides:
$\Rightarrow 3x^2\; dx + 2xy \; dx + x^2\; dy + dx\; y^2 + 2xy \; dy + 3y^2\; dy = 0$
$\Rightarrow (x^2+2xy+3y^2)\; dy = -(3x^2+2xy+y^2)\; dx$
$\Rightarrow \large \frac{dy}{dx}$$= -\large \frac{y^2+2xy+3x^3}{x^2+2xy+3y^2}$