Implicit differentiation (calc 1)

Question

The problem says find $dy/dx$ in terms of $x$ and $y$. I'm having trouble reaching the answer provided on the last line under "should be".

Answer 1

I disagree that the "should-be" answer is correct. Differentiating both sides of the initial equation and solving for $\frac{\mathrm dy}{\mathrm dx}$ gives

$$\begin{align*} \frac{(x+y)\left(1-\frac{\mathrm dy}{\mathrm dx}\right)-(x-y)\left(1+\frac{\mathrm dy}{\mathrm dx}\right)}{(x+y)^2}&=\frac{2xy-x^2\frac{\mathrm dy}{\mathrm dx}}{y^2}\\[1ex] \frac{2y}{(x+y)^2}-\frac{2x}{(x+y)^2}\frac{\mathrm dy}{\mathrm dx}&=\frac{2x}y-\frac{x^2}{y^2}\frac{\mathrm dy}{\mathrm dx}\\[1ex] \left(\frac{x^2}{y^2}-\frac{2x}{(x+y)^2}\right)\frac{\mathrm dy}{\mathrm dx}&=\frac{2x}y-\frac{2y}{(x+y)^2}\\[1ex] \frac{x^2(x+y)^2-2xy^2}{y^2(x+y)^2}\frac{\mathrm dy}{\mathrm dx}&=\frac{2x(x+y)^2-2y^2}{y(x+y)^2}\\[1ex] \frac{\mathrm dy}{\mathrm dx}&=\frac{2xy(x+y)^2-2y^3}{x^2(x+y)^2-2xy^2} \end{align*}$$

which is not the same as the suggested solution.