Let `f :R to R` be a differentiable function satisfying:
`f (xy) =(f(x))/(y)+f(y)/(x) AAx, y in R ^(+) ` also `f (1)=0,f '(1) =1`
find `lim _(x to e) [(1)/(f (x))]` (where [.] denotes greatest integer function).

Answer

To solve the problem, we need to analyze the given function \( f \) and its properties. Let's break down the solution step by step.

### Step 1: Understand the functional equation
We are given the functional equation:
\[ f(xy) = \frac{f(x)}{y} + \frac{f(y)}{x} \]
for all \( x, y \in \mathbb{R}^+ \).

### Step 2: Substitute specific values
Let's substitute \( x = 1 \) into the functional equation:
\[ f(1 \cdot y) = \frac{f(1)}{y} + \frac{f(y)}{1} \]
Since \( f(1) = 0 \), this simplifies to:
\[ f(y) = 0 + f(y) \implies f(y) = f(y) \]
This does not provide new information.

### Step 3: Differentiate the functional equation
Next, we differentiate both sides of the functional equation with respect to \( y \):
\[ \frac{d}{dy} f(xy) = \frac{d}{dy} \left( \frac{f(x)}{y} + \frac{f(y)}{x} \right) \]
Using the chain rule on the left side, we have:
\[ x f'(xy) = -\frac{f(x)}{y^2} + \frac{f'(y)}{x} \]

### Step 4: Evaluate at \( y = 1 \)
Substituting \( y = 1 \) gives:
\[ x f'(x) = -f(x) + \frac{f'(1)}{x} \]
Since \( f'(1) = 1 \), we have:
\[ x f'(x) = -f(x) + \frac{1}{x} \]
Rearranging gives:
\[ x f'(x) + f(x) = \frac{1}{x} \]

### Step 5: Solve the differential equation
This is a first-order linear differential equation. We can rewrite it as:
\[ f'(x) + \frac{f(x)}{x} = \frac{1}{x^2} \]
The integrating factor is \( e^{\int \frac{1}{x} dx} = x \). Multiplying through by the integrating factor:
\[ x f'(x) + f(x) = \frac{1}{x} \]
Integrating both sides:
\[ \int d(x f(x)) = \int \frac{1}{x} dx \]
This gives:
\[ x f(x) = \log x + C \]
Thus:
\[ f(x) = \frac{\log x + C}{x} \]

### Step 6: Use initial conditions to find \( C \)
Using the condition \( f(1) = 0 \):
\[ f(1) = \frac{\log 1 + C}{1} = C = 0 \]
So we have:
\[ f(x) = \frac{\log x}{x} \]

### Step 7: Find \( \lim_{x \to e} \frac{1}{f(x)} \)
Now we need to find:
\[ \lim_{x \to e} \frac{1}{f(x)} = \lim_{x \to e} \frac{x}{\log x} \]
Substituting \( x = e \):
\[ \frac{e}{\log e} = \frac{e}{1} = e \]

### Step 8: Apply the greatest integer function
Finally, we need to find the greatest integer function:
\[ \lfloor e \rfloor \]
Since \( e \approx 2.718 \), we have:
\[ \lfloor e \rfloor = 2 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{2} \]