Let $$y = y(x)$$ be the solution of the differential equation $$\cos x(\log_{e}(\cos x))^{2}dy + (\sin x-3y\sin x\log_{e}(\cos x))dx=0,x \in (0,\frac{\pi}{2})$$. if $$y\left(\frac{\pi}{4}\right) = \frac{-1}{\log_{e}2}$$, then $$y\left(\frac{\pi}{4}\right)$$ is equal to :

$$\cos x(\log\cos x)^2dy+(\sin x-3y\sin x\log\cos x),dx=0$$

Divide by ($$\cos x(\log\cos x)^2):$$
$$\frac{dy}{dx}-\frac{3\tan x}{\log\cos x}y=-\frac{\tan x}{(\log\cos x)^2}$$

Linear DE.

Integrating factor:
$$IF=e^{\int-\frac{3\tan x}{\log\cos x}dx}$$
$$=(\log\cos x)^3$$

$$\frac{d}{dx}\left[y(\log\cos x)^3\right]$$
$$=-\tan x(\log\cos x)$$

Integrate:
$$y(\log\cos x)^3=\frac{(\log\cos x)^2}{2}+C$$

$$y=\frac{1}{2\log\cos x}+\frac{C}{(\log\cos x)^3}$$

Using
$$y\left(\frac{\pi}{4}\right)=-\frac{1}{\log2}$$

gives (C=0).

Hence
$$y=\frac{1}{2\log\cos x}$$

At $$(x=\frac{\pi}{3}),$$
$$\cos\frac{\pi}{3}=\frac{1}{2},\quad\log\frac{1}{2}=-\log2$$

$$y\left(\frac{\pi}{3}\right)=\frac{1}{2(-\log2)}$$
$$=-\frac{1}{2\log2}$$
$$=\frac{1}{\log3-\log4}$$