Let's solve the problem step by step to find where the given function attains its local maxima.

The function is defined as:

\(f(x) = \int_{e^x}^{e^{2x}} t \cdot \log_e t \, \mathrm{dt}\)

To find the extrema, we'll need to compute \(f'(x)\) using the Fundamental Theorem of Calculus and Leibniz Rule. By the Fundamental Theorem,

\(f'(x) = \frac{d}{dx} \left( \int_{e^x}^{e^{2x}} t \cdot \log_e t \, \mathrm{dt} \right).\)

Using the Leibniz Rule for differentiation under the integral sign, the derivative is:

\(f'(x) = \left( e^{2x} \cdot \log_e(e^{2x}) \cdot 2e^{2x} \right) - \left( e^x \cdot \log_e(e^x) \cdot e^x \right).\)

This simplifies to:

\(f'(x) = e^{3x} \cdot 2x - e^{2x} \cdot x.\)

Setting \(f'(x) = 0\) to find critical points:

\(e^{2x} \cdot x(2e^{x} - 1) = 0.\)

Since \(e^{2x} \neq 0\), it must be that

\(x(2e^{x} - 1) = 0.\)

This gives the solutions:

1. \(x = 0\)
2. \(2e^{x} - 1 = 0 \Rightarrow e^{x} = \frac{1}{2} \Rightarrow x = -\log_e 2\)

We know from the problem statement that \(x = 0\) is the point of absolute minima. Thus, we need to verify if \(x = -\log_e 2\) is a point of local maxima by checking the nature of \(f'(x)\) around it.

Consider the sign of \(f'(x)\):

1. For \(x \lt -\log_e 2\), \(2e^{x} \lt 1\); thus, \(f'(x) \gt 0\).
2. For \(x \gt -\log_e 2\), \(2e^{x} \gt 1\); thus, \(f'(x) \lt 0\).

This confirms a change from positive to negative, establishing \(x = -\log_e 2\) as a point of local maxima.

Therefore, the correct answer is:

\(-\log_e 2\)