Answer
$\frac{dy}{dx}=\frac{x-y}{2+x-y}$
$\frac{d^2y}{dx^2}=\frac{4}{(2+x-y)^3}=\frac{1}{2(1+\sqrt y)^3}$
Work Step by Step
Step 1. Given the equation $2\sqrt y=x-y$, we have $2(\frac{1}{2\sqrt y})\frac{dy}{dx}=1-\frac{dy}{dx}$, which gives
$(\frac{1}{\sqrt y}+1)\frac{dy}{dx}=1$; thus $(\frac{2}{x-y}+1)\frac{dy}{dx}=1$, $(\frac{2+x-y}{x-y})\frac{dy}{dx}=1$ and $y'=\frac{dy}{dx}=\frac{x-y}{2+x-y}$
Step 2. Differentiate again with respect to $x$: $y''=\frac{d^2y}{dx^2}=\frac{(2+x-y)(1-y')-(x-y)(1-y')}{(2+x-y)^2}=\frac{2(1-y')}{(2+x-y)^2}=\frac{2((1-\frac{x-y}{2+x-y})}{(2+x-y)^2}=\frac{2(2+x-y-x+y)}{(2+x-y)^3}=\frac{4}{(2+x-y)^3}=\frac{4}{(2+2\sqrt y)^3}=\frac{1}{2(1+\sqrt y)^3}$
