Find all continuous functions $f(x)$ defined for all $x>0$ such that for every $x$, $y > 0$ 
\[ f\left(x+{1\over x}\right)+f\left(y+{1\over y}\right)= f\left(x+{1\over y}\right)+f\left(y+{1\over x}\right) . \]


[i]Proposed by F. Petrov[/i]
We are given the functional equation for a continuous function \( f(x) \) defined for all \( x > 0 \):
\[ f\left(x + \frac{1}{x}\right) + f\left(y + \frac{1}{y}\right) = f\left(x + \frac{1}{y}\right) + f\left(y + \frac{1}{x}\right) \]
for every \( x, y > 0 \).

1. **Substitution and Symmetry:**
   Replace \( x \) by \( \frac{1}{x} \) in the given equation:
   \[ f\left(\frac{1}{x} + x\right) + f\left(y + \frac{1}{y}\right) = f\left(\frac{1}{x} + y\right) + f\left(x + \frac{1}{y}\right) \]
   Since \( \frac{1}{x} + x = x + \frac{1}{x} \), the equation remains unchanged:
   \[ f\left(x + \frac{1}{x}\right) + f\left(y + \frac{1}{y}\right) = f\left(x + y\right) + f\left(\frac{1}{x} + \frac{1}{y}\right) \]

2. **Sequence Definition:**
   Define sequences \( \{a_n\}_{n=0}^\infty \) and \( \{b_n\}_{n=0}^\infty \) as follows:
   \[ a_0 = a, \quad b_0 = b \]
   For \( n \geq 0 \), choose positive reals \( x \) and \( y \) such that:
   \[ x + y = a_n, \quad \frac{1}{x} + \frac{1}{y} = b_n \]
   Define:
   \[ a_{n+1} = x + \frac{1}{x}, \quad b_{n+1} = y + \frac{1}{y} \]

3. **Invariance of the Sum:**
   From the functional equation, we have:
   \[ f(a_0) + f(b_0) = f(a_1) + f(b_1) = \cdots = f(a_n) + f(b_n) = \cdots \quad (\star) \]
   Also, note that \( a_n + b_n \) is a constant throughout.

4. **Convergence Analysis:**
   Consider the numbers \( a_n, b_n, a_{n+1}, b_{n+1} \). Let \( x, y \) be the corresponding values that satisfy \( x + y = a_n \) and \( \frac{1}{x} + \frac{1}{y} = b_n \). This implies \( xy = \frac{a_n}{b_n} \). Then:
   \[
   \frac{|a_{n+1} - b_{n+1}|}{|a_n - b_n|} = \left| \frac{x + \frac{1}{x} - y - \frac{1}{y}}{x + y - \frac{1}{x} - \frac{1}{y}} \right| = \left| \frac{x - y}{x + y} \right| = \sqrt{1 - \frac{4xy}{(x + y)^2}} = \sqrt{1 - \frac{4}{a_n b_n}}
   \]
   Since \( (a_n + b_n)^2 \geq 4a_n b_n \), we have:
   \[
   1 - \frac{4}{a_n b_n} \leq 1 - \frac{16}{(a_n + b_n)^2}
   \]
   Therefore:
   \[
   \frac{|a_{n+1} - b_{n+1}|}{|a_n - b_n|} \leq \sqrt{1 - \frac{16}{(a_n + b_n)^2}} = \sqrt{1 - \frac{16}{(a_0 + b_0)^2}}
   \]
   Since \( \sqrt{1 - \frac{16}{(a_0 + b_0)^2}} \) is a constant less than 1, \( |a_n - b_n| \) converges to zero. Thus, \( a_n \) and \( b_n \) converge to \( \frac{a + b}{2} \).

5. **Functional Equation Simplification:**
   Using the continuity of \( f \) and the fact that \( f(a_0) + f(b_0) = f(a_1) + f(b_1) = \cdots \), we get:
   \[
   f(a) + f(b) = 2f\left(\frac{a + b}{2}\right) \quad \forall a, b > 0 \text{ such that } ab \geq 4 \quad (\spadesuit)
   \]

6. **Generalization:**
   For arbitrary \( a, b > 0 \), choose sufficiently large \( c, d \) such that:
   - \( cd \geq 4 \)
   - \( bc \geq 4 \)
   - \( ad \geq 4 \)
   - \( \frac{a + d}{2} \cdot \frac{b + c}{2} \geq 4 \)
   - \( \frac{a + b}{2} \cdot \frac{c + d}{2} \geq 4 \)

   Applying \( (\spadesuit) \) multiple times, we have:
   \[
   f(a) + f(b) + 2f\left(\frac{c + d}{2}\right) = f(a) + f(b) + f(c) + f(d) = 2f\left(\frac{a + d}{2}\right) + 2f\left(\frac{b + c}{2}\right) = 4f\left(\frac{a + b + c + d}{2}\right) = 2f\left(\frac{a + b}{2}\right) + 2f\left(\frac{c + d}{2}\right)
   \]
   Therefore:
   \[
   f(a) + f(b) = 2f\left(\frac{a + b}{2}\right) \quad \forall a, b > 0
   \]
   By Jensen's inequality, this implies that \( f(x) \) is linear. Hence, \( f(x) = cx + d \) for some constants \( c \) and \( d \).

The final answer is \( \boxed{ f(x) = cx + d } \).