For the curve 2y = y^x+x^y The value of $\frac{\frac{d^{2}y}{d{x}^2}}{\frac{d^{2}x}{d{y}^2}}$at x = 1 is α then |α| =

$$\begin{gathered} \begin{array}{r}2\mathrm{y}={\mathrm{x}}^y+y^x\\ x=1\to 2\mathrm{y}=1+1\to 2\mathrm{y}=2\Rightarrow \mathrm{y}=1\\ 2{\mathrm{y}}^1=x^y\left(\frac{y}{x}+\log \mathrm{x}.{\mathrm{y}}^1\right)+y^x\left(\frac{x{y}^1}{y}+\log \mathrm{y}\right)\end{array} \\ \text{ When }x=1,y=1\Rightarrow 2y^1=(1+0)+1\left(y^1+0\right)\Rightarrow y^1=1 \\ \frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}\Rightarrow \text{ Differentiate with respect to }\mathrm{x} \\ \begin{array}{r}\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\frac{1}{\mathrm{dy}}\Rightarrow \frac{\mathrm{d}}{\mathrm{dy}}\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}\cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\\ \frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dy}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\Rightarrow \frac{\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}}{\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dy}}^2}}=-{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^3=-1\end{array} \end{gathered}$$