Basic Implicit Differentiation Problem

Question

QUESTION: Find $\frac{dy}{dx}$ by implicit differentiation.

$x^2-4xy+y^2=4$

MY SOLUTION

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

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

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

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

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

CORRECT ANSWER

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

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!

Answer 1

The third line should be $$2x - 4y - 4x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$$ not $$2x - 4y\color{red}{\frac{dy}{dx}} \color{red}{+} 4x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$$ and that plus sign error was also in the second line.