Answer
For $x^2-4xy +y^2=4$, find $\frac{dy}{dx}$ by implicit differentiation
$\frac{dy}{dx}=\frac{2y-x}{x-2y}$
Work Step by Step
Differentiate both sides with respect to $x$.
$2x -4(y+x\frac{dy}{dx}) +2y\frac{dy}{dx} = 0$
Isolate $\frac{dy}{dx}$
$2x-4y-4x\frac{dy}{dx} +2y\frac{dy}{dx} =0$
$\frac{dy}{dx}(2y-4x) = 4y-2x$
$\frac{dy}{dx}=\frac{4y-2x}{2y-4x}$
$\frac{dy}{dx}=\frac{2y-x}{x-2y}$