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

$\frac{2}{c^2}$

We are given $\sqrt{x+y}+\sqrt{y-x}=c$               . . . (1)

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

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

$\Rightarrow \sqrt{x+y}-\sqrt{y-x}=\frac{2 x}{c}$         [By (1)]   . . . (2)

(1) + (2) gives      $2 \sqrt{x+y}=c+\frac{2 x}{c}$

Squaring          $4(x+y)=c^2+\frac{4 x^2}{c^2}+4 x$

Canceling 4x,      $4 y=c^2+\frac{4 x^2}{c^2}$

$\Rightarrow 4 \frac{d y}{d x}=\frac{8 x}{c^2} \Rightarrow \frac{d^2 y}{d x^2}=\frac{2}{c^2}$

Hence (3) is correct answer.