The equation of the curve passing through $\left(\frac{\mathrm{\pi }}{6},0\right)$ and satisfying the differential equation $({\mathrm{e}}^{\mathrm{y}}+1)\cos \mathrm{xdx}+{\mathrm{e}}^{\mathrm{y}} \sin \mathrm{xdy}=0,$

$\begin{array}{l}Given(e^y+1)\cos xdx+e^y\sin xdy=0\\ \Rightarrow \frac{\cos x}{\sin x}dx+\frac{e^y}{e^y+1}dy=0\\ \Rightarrow \int \cot xdx+\int \frac{e^y}{e^y+1}dy=\int 0\\ \Rightarrow \log \left|\sin x\right|+\log |e^y+1|=c\\ Itpassesthrough(\frac{\pi }{6},0)\\ \Rightarrow \log \left(\frac{1}{2}\right)+\log 2=c\\ \Rightarrow c=\log 1=0\end{array}$

$\begin{array}{l}\therefore \log \left(\sin x\right)+\log (e^y+1)=0\\ \Rightarrow \log \left(\sin x(e^y+1)\right)=0\\ \Rightarrow \sin x(e^y+1)=1\\ \Rightarrow e^y+1=\cos ecx\\ \Rightarrow e^y=\cos ecx-1\\ \Rightarrow y={\log }_e(\cos ecx-1)\end{array}$