Find all functions $f : R \to R$ satisfying the following conditions
(a) $f(1) = 1$,
(b)  $f(x + y) = f(x) + f(y)$,   $\forall (x,y) \in  R^2$
(c)  $f\left(\frac{1}{x}\right) =\frac{ f(x)}{x^2 }$, $\forall  x \in R -\{0\}$

Trần Nam Dũng	

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   We need to verify that \( f(x) = x \) satisfies all the given conditions:
   - Condition (a): \( f(1) = 1 \) is satisfied since \( f(1) = 1 \).
   - Condition (b): \( f(x + y) = f(x) + f(y) \) is satisfied since \( f(x + y) = x + y \) and \( f(x) + f(y) = x + y \).
   - Condition (c): \( f\left(\frac{1}{x}\right) = \frac{f(x)}{x^2} \) is satisfied since \( f\left(\frac{1}{x}\right) = \frac{1}{x} \) and \( \frac{f(x)}{x^2} = \frac{x}{x^2} = \frac{1}{x} \).

Thus, the only function that satisfies all the given conditions is \( f(x) = x \).

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