Recall the differentiation rule for a logarithm with an arbitrary base.

If $y = \log_b(f(x))$, then $\frac{dy}{dx} = \frac{1}{f(x) \ln b} \cdot f'(x)$.

Here, $f(x) = 3x-2$.
The derivative of $f(x)$ is $f'(x) = \frac{d}{dx}(3x-2) = 3$.

The derivative of $y = \log_b (3x-2)$ is $\boxed{\frac{3}{(3x-2) \ln b}}$.

Since $\log_e x = \ln x$, the function can be written as $y = \ln(\tan x)$.

If $y = \ln(f(x))$, then $\frac{dy}{dx} = \frac{1}{f(x)} \cdot f'(x)$.

Here, $f(x) = \tan x$.
The derivative of $f(x)$ is $f'(x) = \frac{d}{dx}(\tan x) = \sec^2 x$.

Recall that $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$.

This can also be written as $\csc x \sec x$.
The derivative of $y = \log_e (\tan x)$ is $\boxed{\frac{1}{\sin x \cos x}}$.