Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that
                         $f(f(x)f(1-x))=f(x)$    and     $f(f(x))=1-f(x)$,
for all real $x$.
1. **Define the functional equations:**
   We are given two functional equations:
   \[
   f(f(x)f(1-x)) = f(x) \quad \text{(Equation 1)}
   \]
   \[
   f(f(x)) = 1 - f(x) \quad \text{(Equation 2)}
   \]

2. **Identify fixed points:**
   A fixed point of \( f \) is a value \( c \) such that \( f(c) = c \). From Equation 2, we have:
   \[
   f(f(x)) = 1 - f(x)
   \]
   If \( c \) is a fixed point, then \( f(c) = c \). Substituting \( c \) into Equation 2, we get:
   \[
   f(f(c)) = 1 - f(c) \implies f(c) = 1 - c
   \]
   Since \( f(c) = c \), we have:
   \[
   c = 1 - c \implies 2c = 1 \implies c = \frac{1}{2}
   \]
   Therefore, \( \frac{1}{2} \) is a fixed point of \( f \).

3. **Symmetry of the function:**
   Substitute \( x \) with \( 1 - x \) in Equation 1:
   \[
   f(f(1-x)f(x)) = f(1-x)
   \]
   Since \( f(1-x) = f(x) \) (as shown in the solution), we have:
   \[
   f(f(x)f(1-x)) = f(x)
   \]
   This confirms that \( f(1-x) = f(x) \), indicating that \( f \) is symmetric about \( x = \frac{1}{2} \).

4. **Evaluate \( f(f(x)) \):**
   From Equation 2, we have:
   \[
   f(f(x)) = 1 - f(x)
   \]
   Applying \( f \) again, we get:
   \[
   f(f(f(x))) = f(1 - f(x))
   \]
   Since \( f \) is symmetric, \( f(1 - f(x)) = f(f(x)) \). Therefore:
   \[
   f(f(f(x))) = f(f(x))
   \]
   Given that \( f(f(x)) = 1 - f(x) \), we substitute:
   \[
   f(f(x)) = \frac{1}{2}
   \]
   This implies that for all \( x \), \( f(x) = \frac{1}{2} \).

5. **Verification:**
   Substitute \( f(x) = \frac{1}{2} \) into both equations to verify:
   \[
   f(f(x)f(1-x)) = f\left(\frac{1}{2} \cdot \frac{1}{2}\right) = f\left(\frac{1}{4}\right) = \frac{1}{2}
   \]
   \[
   f(x) = \frac{1}{2}
   \]
   \[
   f(f(x)) = f\left(\frac{1}{2}\right) = \frac{1}{2}
   \]
   \[
   1 - f(x) = 1 - \frac{1}{2} = \frac{1}{2}
   \]
   Both equations are satisfied.

Conclusion:
\[
\boxed{f(x) = \frac{1}{2}}
\]