Find all functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying
\[ f\left(\frac {x \plus{} y}{x \minus{} y}\right) \equal{} \frac {f\left(x\right) \plus{} f\left(y\right)}{f\left(x\right) \minus{} f\left(y\right)}
\]
for all $ x \neq y$.
1. **Claim:** \( f(0) = 0 \)

   **Proof:**
   - Consider the assertion \( P(x, 0) \):
     \[
     f\left(\frac{x + 0}{x - 0}\right) = \frac{f(x) + f(0)}{f(x) - f(0)} \implies f(1) = \frac{f(x) + f(0)}{f(x) - f(0)}
     \]
   - Rearranging the equation, we get:
     \[
     f(x) = f(0) \frac{f(1) + 1}{f(1) - 1}
     \]
   - If \( f(0) \neq 0 \), then \( f(x) \) would be a constant function, which does not satisfy the original functional equation. Hence, \( f(0) = 0 \).
     \(\blacksquare\)

2. **Claim:** \( f(1) = 1 \)

   **Proof:**
   - Using the result from the previous claim in \( P(x, 0) \):
     \[
     f(1) = \frac{f(x) + f(0)}{f(x) - f(0)} = \frac{f(x)}{f(x)} = 1
     \]
     \(\blacksquare\)

3. **Claim:** \( f(x) + f(-x) = 0 \) (i.e., \( f \) is odd)

   **Proof:**
   - Consider the assertion \( P(x, -x) \):
     \[
     f\left(\frac{x + (-x)}{x - (-x)}\right) = \frac{f(x) + f(-x)}{f(x) - f(-x)} \implies f(0) = \frac{f(x) + f(-x)}{f(x) - f(-x)}
     \]
   - Since \( f(0) = 0 \), we have:
     \[
     0 = \frac{f(x) + f(-x)}{f(x) - f(-x)} \implies f(x) + f(-x) = 0
     \]
     \(\blacksquare\)

4. **Claim:** For any \( x \in \mathbb{R} \setminus \{0\} \), \( f(x) f\left(\frac{1}{x}\right) = 1 \)

   **Proof:**
   - Consider the assertion \( P(x, -y) \):
     \[
     f\left(\frac{x - y}{x + y}\right) = \frac{f(x) + f(-y)}{f(x) - f(-y)} = \frac{f(x) - f(y)}{f(x) + f(y)}
     \]
   - Combining this with \( P(x, y) \), we get:
     \[
     f\left(\frac{x + y}{x - y}\right) = \frac{f(x) + f(y)}{f(x) - f(y)}
     \]
   - Using the odd property \( f(-y) = -f(y) \), we have:
     \[
     f\left(\frac{x - y}{x + y}\right) = \frac{f(x) - f(y)}{f(x) + f(y)}
     \]
   - From the above, we can deduce that:
     \[
     f(x) f\left(\frac{1}{x}\right) = 1
     \]
     \(\blacksquare\)

5. **Claim:** \( f(xy) = f(x) f(y) \) (i.e., \( f \) is multiplicative)

   **Proof:**
   - Consider the assertion \( P(xy, 1) \):
     \[
     f\left(\frac{xy + 1}{xy - 1}\right) = \frac{f(xy) + f(1)}{f(xy) - f(1)} = \frac{f(xy) + 1}{f(xy) - 1}
     \]
   - Rearranging, we get:
     \[
     f(xy) = \frac{f\left(\frac{xy + 1}{xy - 1}\right) + 1}{f\left(\frac{xy + 1}{xy - 1}\right) - 1}
     \]
   - Using \( P(x, \frac{1}{y}) \):
     \[
     \frac{f\left(\frac{xy + 1}{xy - 1}\right) + 1}{f\left(\frac{xy + 1}{xy - 1}\right) - 1} = f(x) f\left(\frac{1}{y}\right) = f(x) f(y)
     \]
   - Thus, we have:
     \[
     f(xy) = f(x) f(y)
     \]
     \(\blacksquare\)

6. **Claim:** \( f \) is strictly increasing over \( \mathbb{R}^+ \)

   **Proof:**
   - Notice that \( f(x^2) = f(x)^2 \geq 0 \) for all \( x \in \mathbb{R} \), which means \( f(x) > 0 \) for all \( x \in \mathbb{R}^+ \).
   - Suppose we have \( a \) and \( b \) such that \( a > b > 0 \):
     \[
     f\left(\frac{x + y}{x - y}\right) = \frac{f(x) + f(y)}{f(x) - f(y)} > 0
     \]
   - This implies \( f(x) > f(y) \).
     \(\blacksquare\)

7. **Conclusion:**
   - Let \( g(x) = \ln(f(a^x)) \). Notice that \( g(x + y) = g(x) + g(y) \).
   - With all the conditions we have, \( g(x) = cx \), and this means \( f(x) = x^c \).
   - Plugging it into our original equation, we get \( c = 1 \), and thus \( f \equiv x \).

   \(\blacksquare\)

The final answer is \( \boxed{ f(x) = x } \)