$$x^2 - 4xy + y^2 = 4$$
$$2x - 4xy' -4y + 2yy' = 0$$ $$-4xy' + 2yy' = -2x + 4y$$ $$y'(-4 + 2y) = -2x + 4y$$ $$y' = \frac{-2x + 4y}{-4x+2y}$$
But the answer on wolfram is: $\frac{x - 2y}{2x - y}$
Sorry for the newb question. Is this right? And also, can you always multiply the top and bottom of a fraction by $-1$? I guess it makes sense since it's equivalent to $1$?
The two expressions are the same:
$$\frac{-2x+4y}{-4x+2y} = \frac{-2(x-2y)}{-2(2x-y)} = \frac{-2}{-2}\cdot\frac{x-2y}{2x-y} = 1\cdot \frac{x-2y}{2x-y}=\frac{x-2y}{2x-y}$$
You are working with $$x^2 - 4xy + y^2 = 4$$ but you have$$x^2 - 4xy + y = 4$$ in your first line.
Your work is correct assuming your equation is in fact $$x^2 - 4xy + y^2 = 4$$ .
