Let $ E$ be the set of all triangles whose only points with integer coordinates (in the Cartesian coordinate system in space), in its interior or on its sides, are its three vertices, and let $ f$ be the function of area of a triangle. Determine the set of values $ f(E)$ of $ f.$
1. **Understanding the Problem:**
   We need to determine the set of values of the function \( f \) which represents the area of triangles in the set \( E \). The set \( E \) consists of all triangles in 3-dimensional space whose only points with integer coordinates are their vertices.

2. **Analyzing the Given Example:**
   Consider the triangle with vertices \( A(n,0,0) \), \( B(n,1,0) \), and \( C(0,0,1) \). We need to calculate the area of this triangle.

3. **Calculating the Side Lengths:**
   - The length of side \( AB \) is:
     \[
     AB = \sqrt{(n-n)^2 + (1-0)^2 + (0-0)^2} = \sqrt{1} = 1
     \]
   - The length of side \( AC \) is:
     \[
     AC = \sqrt{(n-0)^2 + (0-0)^2 + (0-1)^2} = \sqrt{n^2 + 1}
     \]
   - The length of side \( BC \) is:
     \[
     BC = \sqrt{(n-0)^2 + (1-0)^2 + (0-1)^2} = \sqrt{n^2 + 1 + 1} = \sqrt{n^2 + 2}
     \]

4. **Using the Area Formula for a Triangle in 3D:**
   The area of a triangle with vertices \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \) can be calculated using the following determinant formula:
   \[
   \text{Area} = \frac{1}{2} \sqrt{ \left| \begin{vmatrix}
   x_1 & y_1 & z_1 & 1 \\
   x_2 & y_2 & z_2 & 1 \\
   x_3 & y_3 & z_3 & 1 \\
   \end{vmatrix} \right| }
   \]
   For our triangle \( A(n,0,0) \), \( B(n,1,0) \), and \( C(0,0,1) \), the determinant is:
   \[
   \begin{vmatrix}
   n & 0 & 0 & 1 \\
   n & 1 & 0 & 1 \\
   0 & 0 & 1 & 1 \\
   \end{vmatrix}
   \]
   Expanding this determinant:
   \[
   \begin{vmatrix}
   n & 0 & 0 & 1 \\
   n & 1 & 0 & 1 \\
   0 & 0 & 1 & 1 \\
   \end{vmatrix} = n \begin{vmatrix}
   1 & 0 & 1 \\
   0 & 1 & 1 \\
   \end{vmatrix} - 0 + 0 = n (1 \cdot 1 - 0 \cdot 1) = n
   \]
   Therefore, the area is:
   \[
   \text{Area} = \frac{1}{2} \sqrt{n^2} = \frac{1}{2} n
   \]

5. **Conclusion:**
   Since \( n \) can be any positive integer, the area \( \frac{1}{2} n \) can take any positive value. Therefore, the set of values \( f(E) \) is unbounded and can take any positive real number.

The final answer is \( \boxed{(0, \infty)} \)