16. 已知正实数 $x,y$ 满足 $e^{x}=y\ln x+y\ln y$，则 $\ln y-\dfrac{\ln x+1}{x}$ 的最小值
解：$\because e^{x}=y\ln x+y\ln y$
$\therefore e^{x}=y\ln(xy)$，$xy>1$
$\therefore xe^{x}=xy\ln(xy)=\ln(xy)e^{(\ln(xy))}$
设 $f(x)=xe^{x}$，则 $f(x)=f(\ln(xy))$
$\because f'(x)=(x+1)e^{x}$
$\therefore x>0$ 时 $f(x)$ 单增
$\therefore \ln(xy)=x$
$\therefore y=\dfrac{e^{x}}{x}$
$\therefore \ln y-\dfrac{\ln x+1}{x}=x-\ln x-\dfrac{\ln x+1}{x}$
设 $g(x)=x-\ln x-\dfrac{\ln x+1}{x}$，$g'(x)=\dfrac{x^{2}-x+\ln x}{x^{2}}$
设 $h(x)=x^{2}-x+\ln x$，则 $h'(x)=2x-1+\dfrac{1}{x}=\dfrac{2x^{2}-x+1}{x}=\dfrac{2\left( x-\dfrac{1}{4} \right)^{2}+\dfrac{7}{8}}{x}$
$\therefore x>0$ 时，$h(x)$ 单增
$\because h(1)=0$
$\therefore 0<x<1$ 时，$h(x)<0$，$g'(x)<0$，$g(x)$ 单减
$x>1$ 时，$h(x)>0$，$g'(x)>0$，$g(x)$ 单增
$\therefore g(x)_{极小值}=g(1)=0$
$\therefore \ln y-\dfrac{\ln x+1}{x}$ 的最小值为 $0$