Answer
$\displaystyle \frac{dy}{dx}=\frac{1-y^{2}}{2xy-1}$
Work Step by Step
Differentiating an equation containing
terms of form $f(x)$ and $g(y)$,
terms $f(x)$ are differentiated directly, $\displaystyle \frac{d}{dx}[f(x)],$
and
terms $g(y)$ are differentiated using the chain rule,
$\displaystyle \frac{d}{dx}[g(y)]=\frac{d}{dy}[g(y)]\cdot\frac{dy}{dx}$
------------------
LHS: first term is a product, within which
for $(y^{2})^{\prime}$ we need the chain rule...
$xy^{2}-y=x\displaystyle \qquad/\frac{d}{dx}$
$(1)y^{2}+x(2y\displaystyle \cdot\frac{dy}{dx})-\frac{dy}{dx}=1\qquad/-y^{2}$
$2xy\displaystyle \cdot\frac{dy}{dx}-\frac{dy}{dx}=1-y^{2}$
$\displaystyle \frac{dy}{dx}(2xy-1)=1-y^{2}\qquad/\div(2xy-1)$
$\displaystyle \frac{dy}{dx}=\frac{1-y^{2}}{2xy-1}$
