Answer
$\displaystyle \frac{dy}{dx}=\frac{-2xy}{x^{2}-2y}$
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,
$\displaystyle \frac{d}{dx}[g(y)]=\frac{d}{dy}[g(y)]\cdot\frac{dy}{dx}$
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LHS: sum,
first term: product, ($uv)^{\prime}=u^{\prime}v+uv^{\prime}$
second term: chain rule
RHS: constant
$x^{2}y-y^{2}=4\displaystyle \qquad/\frac{d}{dx}$
$(2x)y+x^{2}\displaystyle \cdot\frac{dy}{dx}-2y\cdot\frac{dy}{dx}=0\qquad/-2xy$
$x^{2}\displaystyle \cdot\frac{dy}{dx}-2y\cdot\frac{dy}{dx}=-2xy$
$\displaystyle \frac{dy}{dx}(x^{2}-2y)=-2xy\qquad/\div(x^{2}-2y)$
$\displaystyle \frac{dy}{dx}=\frac{-2xy}{x^{2}-2y}$
