A real number $\alpha$ is given. Find all functions $f : R^+ \to R^+$ satisfying 
$\alpha x^2f\left(\frac{1}{x}\right) +f(x) =\frac{x}{x+1}$ for all $x > 0$.
1. **Given Functional Equation:**
   \[
   \alpha x^2 f\left(\frac{1}{x}\right) + f(x) = \frac{x}{x+1}
   \]
   for all \( x > 0 \).

2. **Substitute \( x \) with \( \frac{1}{x} \):**
   \[
   \alpha \left(\frac{1}{x}\right)^2 f(x) + f\left(\frac{1}{x}\right) = \frac{\frac{1}{x}}{\frac{1}{x} + 1}
   \]
   Simplify the right-hand side:
   \[
   \frac{\frac{1}{x}}{\frac{1}{x} + 1} = \frac{1}{x + 1}
   \]
   Therefore, the equation becomes:
   \[
   \frac{\alpha}{x^2} f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x+1}
   \]

3. **System of Equations:**
   We now have two equations:
   \[
   \alpha x^2 f\left(\frac{1}{x}\right) + f(x) = \frac{x}{x+1} \quad \text{(1)}
   \]
   \[
   \frac{\alpha}{x^2} f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x+1} \quad \text{(2)}
   \]

4. **Solve the System:**
   Multiply equation (2) by \( x^2 \):
   \[
   \alpha f(x) + x^2 f\left(\frac{1}{x}\right) = \frac{x^2}{x+1}
   \]
   Rearrange to match the form of equation (1):
   \[
   x^2 f\left(\frac{1}{x}\right) + \alpha f(x) = \frac{x^2}{x+1}
   \]
   Compare this with equation (1):
   \[
   \alpha x^2 f\left(\frac{1}{x}\right) + f(x) = \frac{x}{x+1}
   \]
   We now have:
   \[
   x^2 f\left(\frac{1}{x}\right) + \alpha f(x) = \frac{x^2}{x+1} \quad \text{(3)}
   \]
   \[
   \alpha x^2 f\left(\frac{1}{x}\right) + f(x) = \frac{x}{x+1} \quad \text{(4)}
   \]

5. **Equating the Coefficients:**
   From equations (3) and (4), we can equate the coefficients of \( f\left(\frac{1}{x}\right) \) and \( f(x) \):
   \[
   x^2 f\left(\frac{1}{x}\right) + \alpha f(x) = \frac{x^2}{x+1}
   \]
   \[
   \alpha x^2 f\left(\frac{1}{x}\right) + f(x) = \frac{x}{x+1}
   \]

6. **Solving for \( f(x) \):**
   Let \( f(x) = \frac{A x + B}{C x + D} \). Substitute this form into the original equation and solve for \( A, B, C, D \).

   After solving, we find:
   \[
   f(x) = \frac{x - \alpha x^2}{(x+1)(1 - \alpha^2)}
   \]

7. **Verification:**
   Substitute \( f(x) \) back into the original equation to verify that it satisfies the given functional equation.

The final answer is \( \boxed{ f(x) = \frac{x - \alpha x^2}{(x+1)(1 - \alpha^2)} } \).