Given three colors and a rectangle m × n dice unit, we want to color each segment constituting one side of a square drive with one of three colors so that each square unit have two sides of one color and two sides another color. How many colorings we have?
To solve the problem, we need to determine the number of ways to color the edges of an \(m \times n\) rectangle such that each square unit has two sides of one color and two sides of another color. We will use the given lemmas and correct the initial solution.

1. **Isolated Square Coloring:**
   - Each isolated square can be colored in \(18\) ways. This is because we can choose \(2\) colors out of \(3\) in \(\binom{3}{2} = 3\) ways, and for each pair of colors, there are \(6\) valid ways to color the edges (since we can arrange the two colors in \(2\) ways and each arrangement can be rotated in \(3\) ways).
   - Therefore, the number of ways to color an isolated square is \(3 \times 6 = 18\).

2. **Building the \(m \times n\) Rectangle:**
   - We start by coloring the first square, which can be done in \(18\) ways.
   - For the first row, each subsequent square shares one edge with the previous square. According to Lemma 2, each new square can be colored in \(6\) ways given the coloring of the previous square.
   - Therefore, the number of ways to color the first row of \(m\) squares is \(18 \times 6^{m-1}\).

3. **Filling Subsequent Rows:**
   - For each new row, the first square shares one edge with the square directly above it. According to Lemma 2, this can be done in \(6\) ways.
   - Each subsequent square in the row shares two edges (one with the square to the left and one with the square above). According to Lemma 1, each of these squares can be colored in \(2\) ways given the coloring of the previous squares.
   - Therefore, the number of ways to color each new row is \(6 \times 2^{m-1}\).

4. **Combining All Rows:**
   - The total number of ways to color the entire \(m \times n\) rectangle is the product of the number of ways to color the first row and the number of ways to color each subsequent row.
   - The first row can be colored in \(18 \times 6^{m-1}\) ways.
   - Each of the remaining \(n-1\) rows can be colored in \(6 \times 2^{m-1}\) ways.
   - Therefore, the total number of ways to color the \(m \times n\) rectangle is:
     \[
     18 \times 6^{m-1} \times (6 \times 2^{m-1})^{n-1}
     \]
     Simplifying this expression:
     \[
     18 \times 6^{m-1} \times 6^{n-1} \times 2^{(m-1)(n-1)}
     \]
     \[
     18 \times 6^{m+n-2} \times 2^{(m-1)(n-1)}
     \]
     \[
     18 \times 6^{m+n-2} \times 2^{mn-m-n+1}
     \]
     \[
     18 \times 2^{mn-m-n+1} \times 6^{m+n-2}
     \]
     \[
     18 \times 2^{mn-m-n+1} \times (2 \times 3)^{m+n-2}
     \]
     \[
     18 \times 2^{mn-m-n+1} \times 2^{m+n-2} \times 3^{m+n-2}
     \]
     \[
     18 \times 2^{mn-1} \times 3^{m+n-2}
     \]

The final answer is \( \boxed{ 18 \times 2^{mn-1} \times 3^{m+n-2} } \).