Answer
$\displaystyle \frac{dx}{dy}=\frac{3y}{x}$
Work Step by Step
We can find $\displaystyle \frac{dy}{dx}$, and apply $\displaystyle \frac{dx}{dy}=\frac{1}{(\frac{dy}{dx})}.$
$\displaystyle \frac{d}{dx}[x^{2}]=2x$
$\displaystyle \frac{d}{dx}[3y^{2}]=$... chain rule...$=3\displaystyle \cdot 2y\cdot\frac{dy}{dx}=6y\frac{dy}{dx}$
$\displaystyle \frac{d}{dx}[8]=0$
So, after differentiating both sides,
$2x-6y\displaystyle \frac{dy}{dx}=0\qquad /-2x$
$-6y\displaystyle \frac{dy}{dx}=-2x\qquad /\div(-6y)$
$\displaystyle \frac{dy}{dx}=\frac{-2x}{-6y}=\frac{x}{3y}$
$\displaystyle \frac{dx}{dy}=\frac{1}{(\frac{dy}{dx})}=\frac{3y}{x}$
$\displaystyle \frac{dx}{dy}=\frac{3y}{x}$
