A function $f(x)$ satisfies $f(x)=f\left(\frac{c}{x}\right)$  for some real number $c(>1)$ and all positive number $x$. 

If $\int_1^{\sqrt{c}} \frac{f(x)}{x} dx=3$, evaluate $\int_1^c \frac{f(x)}{x} dx$
1. Given the function \( f(x) \) satisfies \( f(x) = f\left(\frac{c}{x}\right) \) for some real number \( c > 1 \) and all positive numbers \( x \).

2. We are given that:
   \[
   \int_1^{\sqrt{c}} \frac{f(x)}{x} \, dx = 3
   \]

3. We need to evaluate:
   \[
   \int_1^c \frac{f(x)}{x} \, dx
   \]

4. We can split the integral from 1 to \( c \) into two parts:
   \[
   \int_1^c \frac{f(x)}{x} \, dx = \int_1^{\sqrt{c}} \frac{f(x)}{x} \, dx + \int_{\sqrt{c}}^c \frac{f(x)}{x} \, dx
   \]

5. We already know that:
   \[
   \int_1^{\sqrt{c}} \frac{f(x)}{x} \, dx = 3
   \]

6. Now, consider the second part of the integral:
   \[
   \int_{\sqrt{c}}^c \frac{f(x)}{x} \, dx
   \]

7. Using the substitution \( u = \frac{c}{x} \), we get \( du = -\frac{c}{x^2} \, dx \) or \( dx = -\frac{x^2}{c} \, du \). When \( x = \sqrt{c} \), \( u = \sqrt{c} \). When \( x = c \), \( u = 1 \).

8. Therefore:
   \[
   \int_{\sqrt{c}}^c \frac{f(x)}{x} \, dx = \int_{\sqrt{c}}^1 \frac{f\left(\frac{c}{u}\right)}{\frac{c}{u}} \left(-\frac{u^2}{c}\right) \, du
   \]

9. Simplifying the integrand:
   \[
   \int_{\sqrt{c}}^1 \frac{f\left(\frac{c}{u}\right)}{\frac{c}{u}} \left(-\frac{u^2}{c}\right) \, du = \int_{\sqrt{c}}^1 \frac{f(u)}{u} \, du
   \]

10. Changing the limits of integration:
    \[
    \int_{\sqrt{c}}^1 \frac{f(u)}{u} \, du = -\int_1^{\sqrt{c}} \frac{f(u)}{u} \, du
    \]

11. Therefore:
    \[
    \int_{\sqrt{c}}^c \frac{f(x)}{x} \, dx = -\int_1^{\sqrt{c}} \frac{f(u)}{u} \, du
    \]

12. Given that:
    \[
    \int_1^{\sqrt{c}} \frac{f(x)}{x} \, dx = 3
    \]

13. We have:
    \[
    \int_{\sqrt{c}}^c \frac{f(x)}{x} \, dx = -3
    \]

14. Combining the two parts:
    \[
    \int_1^c \frac{f(x)}{x} \, dx = \int_1^{\sqrt{c}} \frac{f(x)}{x} \, dx + \int_{\sqrt{c}}^c \frac{f(x)}{x} \, dx = 3 + (-3) = 0
    \]

The final answer is \(\boxed{0}\)