If $\sqrt{x+y}+\sqrt{y-x}=c,\text{then}\frac{d^{2}y}{d{x}^2}=$ 

$\text{Consider}(\sqrt{x+y}{)}^2-(\sqrt{y-x}{)}^2=$

$(\sqrt{x+y}-\sqrt{y-x})(\sqrt{x+y}+\sqrt{y+x})$

$\Rightarrow 2x=(\sqrt{x+y}-\sqrt{y-x})\cdot c$

$\Rightarrow \sqrt{x+y}-\sqrt{y-x}=\frac{2x}{c}\to \left(1\right)$

$\Rightarrow \text{given}\sqrt{x+y}+\sqrt{y-x}=c\to \left(2\right)$

$\left(1\right)+\left(2\right):2\sqrt{x+y}=\frac{2x}{c}+c$

$\text{ Squaring both sides, we get }4x+4y=c^2+4x+\frac{4x^2}{c^2}$

$\begin{array}{r}\therefore y=\frac{c^2}{4}+\frac{x^2}{c^2}\therefore \frac{dy}{dx}=\frac{2x}{c^2}\\ \therefore \frac{d^{2}y}{d{x}^2}=\frac{2}{c^2}\end{array}$