Consider `f: R^+vecRs u c ht h a tf(3)=1` for `a in R^+a n df(x)dotf(y)+f(3/x)f(3/y)=2f(x y)AAx ,y in R^+dot` Then find `f(x)dot`

Explanation

To solve the problem, we need to find the function \( f(x) \) given the functional equation:

\[ f(x) \cdot f(y) + f\left(\frac{3}{x}\right) \cdot f\left(\frac{3}{y}\right) = 2 f(xy) \]

with the condition that \( f(3) = 1 \).

### Step-by-Step Solution:

1. **Substituting Values**:
Let's first substitute \( x = 1 \) and \( y = 1 \) into the functional equation.
\[ f(1) \cdot f(1) + f(3) \cdot f(3) = 2 f(1 \cdot 1) \]
This simplifies to:
\[ f(1)^2 + 1^2 = 2 f(1) \]
Thus, we have:
\[ f(1)^2 + 1 = 2 f(1) \]

2. **Rearranging the Equation**:
Rearranging gives us:
\[ f(1)^2 - 2f(1) + 1 = 0 \]
This can be factored as:
\[ (f(1) - 1)^2 = 0 \]
Therefore, we find:
\[ f(1) = 1 \]

3. **Substituting Again**:
Now, let’s substitute \( y = 1 \) and keep \( x \) as \( x \) in the original equation:
\[ f(x) \cdot f(1) + f\left(\frac{3}{x}\right) \cdot f(3) = 2 f(x \cdot 1) \]
Since \( f(1) = 1 \) and \( f(3) = 1 \), this simplifies to:
\[ f(x) + f\left(\frac{3}{x}\right) = 2f(x) \]
Rearranging gives:
\[ f\left(\frac{3}{x}\right) = f(x) \]

4. **Using the New Relation**:
Now, we can substitute \( y = \frac{3}{x} \) into the original equation:
\[ f(x) \cdot f\left(\frac{3}{x}\right) + f(3) \cdot f(3) = 2 f\left(x \cdot \frac{3}{x}\right) \]
This simplifies to:
\[ f(x) \cdot f(x) + 1 = 2 f(3) \]
Since \( f(3) = 1 \), we have:
\[ f(x)^2 + 1 = 2 \]
Thus:
\[ f(x)^2 = 1 \]

5. **Finding the Function**:
Taking the square root gives us:
\[ f(x) = 1 \quad \text{or} \quad f(x) = -1 \]
However, since \( f: \mathbb{R}^+ \to \mathbb{R} \) and must be positive, we conclude:
\[ f(x) = 1 \]

### Final Answer:
Thus, the function \( f(x) \) is:
\[ \boxed{1} \]