If $x^x{y}^y=e^e$, then ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{(e,e)}=$

$x^x{y}^y=e^e$

Apply log on sides

$x\log x+y\log y=e\log e$

Diff with respect to x

$1+\log x+\frac{dy}{dx}+\log y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-\left(1+\log x\right)}{\left(1+\log y\right)}$

$\frac{d^{2}y}{d{x}^2}=\frac{{\left({\displaystyle \frac{dy}{dx}}\right)}^2\left({\displaystyle \frac{1}{y}}\right)-\left({\displaystyle \frac{1}{x}}\right)}{\left(1+\log y\right)}$