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Solution

Let $a=\sqrt{x+y},\,b=\sqrt{y-x}$. $(a+b)^2=2 \Rightarrow y+\sqrt{y^{2}-x^{2}}=1$. Differentiate: $y'+\dfrac{yy'-x}{\sqrt{y^{2}-x^{2}}}=0$. But $\sqrt{y^{2}-x^{2}}=1-y$ from above ⇒ $y'=x$ ⇒ $y''=1$.