${\text{If e}}^{x+y}+e^{y-x}=1, then the value of y^{"}-{\left(y^{\text{'}}\right)}^{2 }is$

$\begin{array}{l}{\text{e}}^{x+y}+e^{y-x}=1\\ \text{}\Rightarrow e^x+e^{-x} =e^{-y}\text{} \\ Differentiating w.r.t. x, we get\\ \text{}{e}^x-e^{-x} =-e^{-y}\frac{dy}{dx}\text{}\\ Again differentiating w.r.t. x, we get\\ \text{}{e}^x+e^{-x} =-e^{-y}\frac{d^{2}y}{d{x}^2}+e^{-y}{\left(\frac{dy}{dx}\right)}^2 \text{}\frac{d^{2}y}{d{x}^2}-{\left(\frac{dy}{dx}\right)}^2\\ =-e^y\left(e^x+e^{-x} \right)=-1\end{array}$