Answer
$\displaystyle \frac{dx}{dy}=-\frac{x+1}{xy}$
Work Step by Step
LHS, first term: chain rule,
$\displaystyle \frac{d}{dy}[(xy)^{2}]=2(xy)\cdot\frac{d}{dy}[xy]$
$\displaystyle \frac{d}{dy}[xy]$=...product...$=\displaystyle \frac{dx}{dy}\cdot y+1$
$\displaystyle \frac{d}{dy}[(xy)^{2}]=2(xy)\cdot(\frac{dx}{dy}\cdot y+1)$
$LHS$, second term,
$\displaystyle \frac{d}{dy}[y^{2}]=2y$
RHS, $\displaystyle \frac{d}{dy}[8]=0$
So, after differentiating both sides,
we solve for $\displaystyle \frac{dx}{dy}$:
$2(xy)\displaystyle \cdot(\frac{dx}{dy}\cdot y+1)+2y=0$
$\displaystyle \frac{dx}{dy}\cdot 2xy^{2}+2xy+2y=0\qquad/-2xy-2y$
$\displaystyle \frac{dx}{dy}\cdot 2xy^{2}=-2xy-2y\qquad/\div 2xy^{2}$
$\displaystyle \frac{dx}{dy}=\frac{-2y(x+1)}{2xy^{2}}=-\frac{x+1}{xy}$
