Finding the dy/dx of a complicated function

Question

I need urgent help on this question. I have no clue how to solve it as it's very complicated to me. The question is the following:

Given $y=\frac{2xy}{x^2 + y}$ find $\frac{dy}{dx}$.

Answer 1

$$y=\frac{2xy}{x^2+y}\implies x^2+y=2x\implies y=2x-x^2~\forall~y\neq 0$$

I hope you can take it from here.

Answer 2

Differentiate both sides w.r.t. x (by applying quotient rule):

\begin{align} y' & =\frac{(2y+2xy')(x^2+y)-2xy(2x+y')}{(x^2+y)^2}\\ &=\frac{2x^2y+2x^3y'+2y^2+2xyy' -4x^2y -2xyy'}{(x^2+y)^2}\\ &= \frac{2x^3y'+2y^2-2x^2y}{(x^2+y)^2}\\ (x^2+y)^2y' &= 2x^3y'+2y^2 -2x^2y\\ y' &= \frac{2y^2 -2x^2y}{(x^2+y)^2-2x^3} \end{align}

Answer 3

You can cancel y in numerator on each side and express $ y=f(x) $

$ y = 2 x - x^2 $

$ \dfrac{dy}{dx} = 2 - 2 x $