1. **State the problem:** Find the length of the curve defined by $y = \log_e(\sec x)$ for $0 \leq x \leq \frac{\pi}{3}$.\n\n2. **Formula for arc length:** The length $L$ of a curve $y = f(x)$ from $x=a$ to $x=b$ is given by:\n$$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$\n\n3. **Find the derivative $\frac{dy}{dx}$:**\nGiven $y = \log_e(\sec x)$, use the chain rule:\n$$\frac{dy}{dx} = \frac{1}{\sec x} \cdot \frac{d}{dx}(\sec x)$$\nRecall $\frac{d}{dx}(\sec x) = \sec x \tan x$, so:\n$$\frac{dy}{dx} = \frac{1}{\sec x} \cdot \sec x \tan x = \tan x$$\n\n4. **Substitute into the arc length formula:**\n$$L = \int_0^{\frac{\pi}{3}} \sqrt{1 + (\tan x)^2} \, dx$$\nRecall the identity $1 + \tan^2 x = \sec^2 x$, so:\n$$L = \int_0^{\frac{\pi}{3}} \sqrt{\sec^2 x} \, dx = \int_0^{\frac{\pi}{3}} \sec x \, dx$$\n\n5. **Evaluate the integral:**\nThe integral of $\sec x$ is:\n$$\int \sec x \, dx = \ln |\sec x + \tan x| + C$$\nSo,\n$$L = \left[ \ln |\sec x + \tan x| \right]_0^{\frac{\pi}{3}} = \ln(\sec \frac{\pi}{3} + \tan \frac{\pi}{3}) - \ln(\sec 0 + \tan 0)$$\n\n6. **Calculate values:**\n$\sec \frac{\pi}{3} = 2$, $\tan \frac{\pi}{3} = \sqrt{3}$, $\sec 0 = 1$, $\tan 0 = 0$.\n\n7. **Final length:**\n$$L = \ln(2 + \sqrt{3}) - \ln(1) = \ln(2 + \sqrt{3})$$\n\n**Answer:** The length of the curve is $\boxed{\ln(2 + \sqrt{3})}$.