Answer
$\displaystyle \frac{dy}{dx}=-\frac{6+9x^{2}y}{9x^{3}-x^{2}}$
Work Step by Step
$\displaystyle \frac{d}{dx}[3xy]=$..product...$= 3\displaystyle \cdot[1\cdot y+x\cdot\frac{dy}{dx}]$
$=3y+3x\displaystyle \cdot\frac{dy}{dx}$
$\displaystyle \frac{d}{dx}[\frac{y}{3}]=\frac{1}{3}\cdot\frac{dy}{dx}$
$\displaystyle \frac{d}{dx}[\frac{2}{x}]=\frac{d}{dx}[2x^{-1}]=2(-x^{-2})=-\frac{2}{x^{2}}$
so, after differentiating both sides,
$3y+3x\displaystyle \cdot\frac{dy}{dx}-\frac{1}{3}\cdot\frac{dy}{dx}=-\frac{2}{x^{2}}$
which we solve for $\displaystyle \frac{dy}{dx}$
$\displaystyle \frac{dy}{dx}(3x-\frac{1}{3})=-\frac{2}{x^{2}}-3y$
$\displaystyle \frac{dy}{dx}=\frac{-\frac{2}{x^{2}}-3y}{(3x-\frac{1}{3})}=\frac{-2-3x^{2}y}{x^{2}(3x-\frac{1}{3})}$
$=\displaystyle \frac{-(2+3x^{2}y)}{x^{2}(\frac{9x-1}{3})}$
$=-\displaystyle \frac{3(2+3x^{2}y)}{x^{2}(9x-1)}$
$\displaystyle \frac{dy}{dx}=-\frac{6+9x^{2}y}{9x^{3}-x^{2}}$
