$$\begin{aligned} & \cos x(\ln (\cos x))^2 d y+(\sin x-3 y(\sin x) \ln (\cos x)) d x=0 \\ & \cos x(\ln (\cos x))^2 \frac{d y}{d x}-3 \sin x \cdot \ln (\cos x) y=-\sin x \\ & \frac{d y}{d x}-\frac{3 \tan x}{\ln (\cos x)} y=\frac{-\tan x}{(\ln (\cos x))^2} \\ & \frac{d y}{d x}+\frac{3 \tan x}{\ln (\sec x)} y=\frac{-\tan x}{(\ln (\sec x))^2} \\ & \text { I.F. }=e^{\int \frac{3 \tan x}{\ln (\sec x)} d x}=(\ln (\sec x))^3 \end{aligned}$$

$$\begin{aligned} & y \times(\ln (\sec x))^3=-\int \frac{\tan x}{(\ln (\sec x))^2}(\ln (\sec x))^3 d x \\ & y \times(\ln (\sec x))^3=-\frac{1}{2}(\ln (\sec x))^2+C \\ & \text { Given }: x=\frac{\pi}{4}, y=-\frac{1}{\ln 2} \\ & \frac{-1}{\ln 2} \times(\ln \sqrt{2})^3=-\frac{1}{2} \times(\ln \sqrt{2})^2+C \\ & \Rightarrow \frac{-1}{8 \ln 2} \times(\ln 2)^3=\frac{-1}{2} \times \frac{1}{4}(\ln 2)^2+C \\ & -\frac{1}{8}(\ln 2)^2=\frac{-1}{8}(\ln 2)^2+C \\ & \Rightarrow C=0 \\ & \therefore y(\ln (\sec x))^3=\frac{-1}{2}(\ln (\sec x))^2+0 \\ & y=\frac{-1}{2 \ln (\sec x)} \\ & y=\frac{1}{2 \ln (\cos x)} \end{aligned}$$

$$\begin{aligned} & \therefore y\left(\frac{\pi}{6}\right)=\frac{1}{2 \ln \left(\cos \frac{\pi}{6}\right)} \\ & =\frac{1}{2 \ln \left(\frac{\sqrt{3}}{2}\right)} \\ & =\frac{1}{2\left(\frac{1}{2} \ln 3-\ln 2\right)} \\ & =\frac{1}{\ln 3-\ln 4} \end{aligned}$$