Determine all functions $f : \mathbb{R} - [0,1] \to \mathbb{R}$ such that \[ f(x) + f \left( \dfrac{1}{1-x} \right) = \dfrac{2(1-2x)}{x(1-x)} .  \]
1. Given the functional equation:
   \[
   f(x) + f \left( \dfrac{1}{1-x} \right) = \dfrac{2(1-2x)}{x(1-x)}
   \]
   Let's denote this as equation \((1)\).

2. Substitute \( x \) with \( \frac{1}{1-x} \) in equation \((1)\):
   \[
   f \left( \frac{1}{1-x} \right) + f \left( \frac{x-1}{x} \right) = \dfrac{2 \left( 1 - 2 \cdot \frac{1}{1-x} \right)}{\frac{1}{1-x} \left( 1 - \frac{1}{1-x} \right)}
   \]
   Simplify the right-hand side:
   \[
   \dfrac{2 \left( 1 - \frac{2}{1-x} \right)}{\frac{1}{1-x} \left( \frac{x-1}{1-x} \right)} = \dfrac{2 \left( \frac{1-x-2}{1-x} \right)}{\frac{1}{1-x} \cdot \frac{x-1}{1-x}} = \dfrac{2 \left( \frac{1-2x}{1-x} \right)}{\frac{x-1}{(1-x)^2}} = \dfrac{2(1-2x)}{x(1-x)}
   \]
   Thus, we have:
   \[
   f \left( \frac{1}{1-x} \right) + f \left( \frac{x-1}{x} \right) = \dfrac{2(1-2x)}{x(1-x)}
   \]
   Denote this as equation \((2)\).

3. Substitute \( x \) with \( \frac{x-1}{x} \) in equation \((1)\):
   \[
   f \left( \frac{x-1}{x} \right) + f \left( \frac{1}{1 - \frac{x-1}{x}} \right) = \dfrac{2 \left( 1 - 2 \cdot \frac{x-1}{x} \right)}{\frac{x-1}{x} \left( 1 - \frac{x-1}{x} \right)}
   \]
   Simplify the right-hand side:
   \[
   \dfrac{2 \left( \frac{x - 2(x-1)}{x} \right)}{\frac{x-1}{x} \left( \frac{2x-1}{x} \right)} = \dfrac{2 \left( \frac{x - 2x + 2}{x} \right)}{\frac{x-1}{x} \cdot \frac{2x-1}{x}} = \dfrac{2 \left( \frac{2-x}{x} \right)}{\frac{(x-1)(2x-1)}{x^2}} = \dfrac{2(2-x)x}{(x-1)(2x-1)}
   \]
   Thus, we have:
   \[
   f \left( \frac{x-1}{x} \right) + f(x) = \dfrac{2(2x-1)x}{(x-1)}
   \]
   Denote this as equation \((3)\).

4. Now we have three equations:
   \[
   \begin{cases}
   f(x) + f \left( \dfrac{1}{1-x} \right) = \dfrac{2(1-2x)}{x(1-x)} & \text{(1)} \\
   f \left( \frac{1}{1-x} \right) + f \left( \frac{x-1}{x} \right) = \dfrac{2(1-2x)}{x(1-x)} & \text{(2)} \\
   f \left( \frac{x-1}{x} \right) + f(x) = \dfrac{2(2x-1)x}{(x-1)} & \text{(3)}
   \end{cases}
   \]

5. Solving these equations, we can see that:
   \[
   f(x) = \dfrac{x+1}{x-1}
   \]
   satisfies all three equations.

The final answer is \(\boxed{\dfrac{x+1}{x-1}}\).