Let $y = y(x)$ be the solution of the differential equation $\cos x (\log_e (\cos x))^2 dy + (\sin x - 3y \sin x \log_e (\cos x)) dx = 0, x \in \left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{4}\right) = \frac{-1}{\log_e 2}$, then $y\left(\frac{\pi}{6}\right)$ is equal to:

Step 1: Rearrange the differential equation

First, we need to rearrange the given differential equation into the standard form of a linear first-order differential equation, which is $\frac{dy}{dx} + P(x)y = Q(x)$. This involves dividing by $dx$ and isolating the terms involving $y$ and constants.

Step 2: Identify P(x) and Q(x)

From the rearranged equation, we can identify the functions $P(x)$ and $Q(x)$. These functions are crucial for finding the integrating factor.

Step 3: Calculate the integrating factor

The integrating factor (I.F.) is calculated using the formula $e^{\int P(x) dx}$. We perform a substitution $u = \log_e (\cos x)$, so $du = \frac{1}{\cos x} (-\sin x) dx = -\tan x dx$. This simplifies the integral, leading to the integrating factor $(\log_e (\cos x))^{-3}$.

Step 4: Solve the differential equation

The general solution is given by $y \cdot I.F. = \int Q(x) \cdot I.F. dx$. We substitute the expressions for $Q(x)$ and the I.F. and integrate. Again, using the substitution $u = \log_e (\cos x)$, the integral becomes straightforward, yielding the general solution for $y(x)$.

Step 5: Apply the initial condition to find C

We use the given initial condition $y\left(\frac{\pi}{4}\right) = \frac{-1}{\log_e 2}$ to find the value of the constant $C$. Substituting $x = \frac{\pi}{4}$ into the general solution and using $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, we find that $C=0$.

Step 6: Calculate $y\left(\frac{\pi}{6}\right)$

With $C=0$, the particular solution is $y(x) = \frac{1}{2} (\log_e (\cos x))^{-2}$. Now, we substitute $x = \frac{\pi}{6}$ into this solution. Since $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$, we can calculate the final value of $y\left(\frac{\pi}{6}\right)$.

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