Find dydx by implicit differentiation.

x2−4xy+y2=4

My solution:

$\frac{d}{dx}{x}^{2}-\frac{d}{dx}4xy+\frac{d}{dx}{y}^{2}=\frac{d}{dx}4$

$2x-(4x{)}^{\prime}(y)+(4x)(y{)}^{\prime}+2y\frac{dy}{dx}=0$

$2x-4y\frac{dy}{dx}+4x\frac{dy}{dx}+2y\frac{dy}{dx}=0$

$\frac{dy}{dx}(-4y+4x+2y)=-2x$

$\frac{dy}{dx}=\frac{-2x}{4x-2y}$

Correct answer:

$\frac{dy}{dx}=\frac{2y-x}{y-2x}$

I would like to understand why my answer is incorrect. If someone could take a look at my steps and explain to me where I went wrong it would be greatly appreciated!

x2−4xy+y2=4

My solution:

$\frac{d}{dx}{x}^{2}-\frac{d}{dx}4xy+\frac{d}{dx}{y}^{2}=\frac{d}{dx}4$

$2x-(4x{)}^{\prime}(y)+(4x)(y{)}^{\prime}+2y\frac{dy}{dx}=0$

$2x-4y\frac{dy}{dx}+4x\frac{dy}{dx}+2y\frac{dy}{dx}=0$

$\frac{dy}{dx}(-4y+4x+2y)=-2x$

$\frac{dy}{dx}=\frac{-2x}{4x-2y}$

Correct answer:

$\frac{dy}{dx}=\frac{2y-x}{y-2x}$

I would like to understand why my answer is incorrect. If someone could take a look at my steps and explain to me where I went wrong it would be greatly appreciated!