(D)
Given $\sqrt{x+y}+\sqrt{y-x}=5 \Rightarrow \sqrt{y-x}=5-\sqrt{x+y}$
On squaring both side, we get
$$
y-x=25+x+y-10 \sqrt{x+y} \Rightarrow 10 \sqrt{x+y}-2 x=25
$$
Differentiating w.r.t. $x$,
$10 \times \frac{1}{2 \sqrt{x+y}}\left(1+\frac{d y}{d x}\right)-2 \times 1=0 \Rightarrow \frac{5}{\sqrt{x+y}}\left(1+\frac{d y}{d x}\right)=2$
$$
\therefore 1+\frac{d y}{d x}=\frac{2 \sqrt{x+y}}{5}....(1)
$$
Differentiating w.r.t. $x$,
$$
\begin{aligned}
\frac{d^{2} y}{d x^{2}} &=\frac{2}{5} \times \frac{1}{2 \sqrt{x+y}}\left(1+\frac{d y}{d x}\right)=\frac{1}{5 \sqrt{x+y}} \times \frac{2 \sqrt{x+y}}{5} \quad \ldots[\because \text { From (1) }] \\
&=\frac{2}{25}
\end{aligned}
$$