Find all monotonous functions $ f: \mathbb{R} \to \mathbb{R}$ that satisfy the following functional equation:
\[f(f(x)) \equal{} f( \minus{} f(x)) \equal{} f(x)^2.\]
1. **Given Functional Equation:**
   \[
   f(f(x)) = f(-f(x)) = f(x)^2
   \]
   We need to find all monotonous functions \( f: \mathbb{R} \to \mathbb{R} \) that satisfy this equation.

2. **Symmetry and Monotonicity:**
   Since \( f \) is monotonous, it is either non-decreasing or non-increasing. Let's assume \( f \) is non-decreasing. The same argument will hold for non-increasing functions by symmetry.

3. **Analyzing \( f(f(x)) = f(-f(x)) \):**
   \[
   f(f(x)) = f(-f(x))
   \]
   Since \( f \) is non-decreasing, \( f(x) \) must be constant on any interval where \( f \) is constant. This implies that \( f(x) \) is constant on the range of \( f \).

4. **Constant Function Analysis:**
   Suppose \( f(x) = c \) for all \( x \in \mathbb{R} \). Then:
   \[
   f(f(x)) = f(c) = c
   \]
   and
   \[
   f(-f(x)) = f(-c) = c
   \]
   Therefore, \( c = c^2 \), which gives us two solutions:
   \[
   c = 0 \quad \text{or} \quad c = 1
   \]

5. **Case 1: \( f(x) = 0 \):**
   If \( f(x) = 0 \) for all \( x \in \mathbb{R} \), then:
   \[
   f(f(x)) = f(0) = 0
   \]
   and
   \[
   f(-f(x)) = f(0) = 0
   \]
   This satisfies the given functional equation. Hence, \( f(x) = 0 \) is a solution.

6. **Case 2: \( f(x) = 1 \):**
   If \( f(x) = 1 \) for all \( x \in \mathbb{R} \), then:
   \[
   f(f(x)) = f(1) = 1
   \]
   and
   \[
   f(-f(x)) = f(-1) = 1
   \]
   This satisfies the given functional equation. Hence, \( f(x) = 1 \) is a solution.

7. **Non-constant Solutions:**
   Suppose \( f \) is not constant. Then \( f(x) \) must take on at least two distinct values. Let \( a \) and \( b \) be such that \( f(a) = 1 \) and \( f(b) = -1 \). However, since \( f \) is monotonous and \( f(x)^2 \geq 0 \), \( f(x) \) can only take values in \(\{0, 1\}\).

8. **Monotonicity and Range:**
   Given \( f \) is monotonous and can only take values in \(\{0, 1\}\), we have the following possibilities:
   - \( f(x) = 1 \) for all \( x \geq a \) and \( f(x) = 0 \) for all \( x < a \).
   - \( f(x) = 0 \) for all \( x \geq a \) and \( f(x) = 1 \) for all \( x < a \).

9. **Verification:**
   For \( f(x) = 1 \) for all \( x \geq a \) and \( f(x) = 0 \) for all \( x < a \):
   \[
   f(f(x)) = \begin{cases} 
   f(1) = 1 & \text{if } x \geq a \\
   f(0) = 0 & \text{if } x < a 
   \end{cases}
   \]
   and
   \[
   f(-f(x)) = \begin{cases} 
   f(-1) = 1 & \text{if } x \geq a \\
   f(0) = 0 & \text{if } x < a 
   \end{cases}
   \]
   This does not satisfy the functional equation for all \( x \).

10. **Conclusion:**
    The only solutions are the constant functions \( f(x) = 0 \) and \( f(x) = 1 \).

The final answer is \( \boxed{ f(x) = 0 } \) or \( f(x) = 1 \).