The trinomial $f(x)$ is such that $(f(x))^3-f(x)=0$ has three real roots. Find the y-coordinate of the vertex of $f(x)$.
1. Given the equation \((f(x))^3 - f(x) = 0\), we can factor it as follows:
   \[
   (f(x))^3 - f(x) = f(x) \cdot (f(x)^2 - 1) = f(x) \cdot (f(x) - 1) \cdot (f(x) + 1) = 0
   \]
   This implies that \(f(x) = 0\), \(f(x) - 1 = 0\), or \(f(x) + 1 = 0\).

2. Therefore, the roots of the equation are:
   \[
   f(x) = 0, \quad f(x) = 1, \quad f(x) = -1
   \]
   Since the problem states that there are three real roots, one of these factors must have a double root. This means that one of the roots must be repeated.

3. To determine which root is repeated, we need to consider the nature of the polynomial \(f(x)\). Since \(f(x)\) is a trinomial (a polynomial of degree 2), it can be written in the form:
   \[
   f(x) = ax^2 + bx + c
   \]

4. For \(f(x)\) to have three real roots (considering multiplicity), one of the roots must be a double root. Let's assume \(f(x) = 0\) has a double root. This means the polynomial can be written as:
   \[
   f(x) = a(x - r)^2
   \]
   where \(r\) is the root with multiplicity 2.

5. Since \(f(x) = 0\) has a double root, the other roots \(f(x) = 1\) and \(f(x) = -1\) must be single roots. This implies that the polynomial \(f(x)\) must be symmetric about the y-axis, and the vertex of the parabola \(f(x)\) must be at \(y = 0\).

6. The y-coordinate of the vertex of the polynomial \(f(x)\) is the value of \(f(x)\) at the vertex. Since the vertex is at the origin (0,0), the y-coordinate of the vertex is:
   \[
   \boxed{0}
   \]