$f(x)$ is a continuous function defined in $x>0$. For all $a,\ b\ (a>0,\ b>0)$, if $\int_a^b f(x)\ dx$ is determined by only $\frac{b}{a}$, then prove that $f(x)=\frac{c}{x}\ (c: constant).$
1. Given that \( f(x) \) is a continuous function defined for \( x > 0 \), and for all \( a, b \) with \( a > 0 \) and \( b > 0 \), the integral \( \int_a^b f(x) \, dx \) is determined only by the ratio \( \frac{b}{a} \).

2. Let \( \frac{b}{a} = k \). Then, we can write \( b = ka \).

3. Define \( I(a, b) = \int_a^b f(x) \, dx \). According to the problem, \( I(a, b) \) depends only on \( k \), so we can write \( I(a, b) = g(k) \) for some function \( g \).

4. Consider the integral from \( a \) to \( ka \):
   \[
   I(a, ka) = \int_a^{ka} f(x) \, dx = g(k)
   \]

5. Now, consider the integral from \( ka \) to \( k^2a \):
   \[
   I(ka, k^2a) = \int_{ka}^{k^2a} f(x) \, dx = g(k)
   \]

6. By the additivity property of definite integrals, we have:
   \[
   I(a, k^2a) = I(a, ka) + I(ka, k^2a)
   \]
   Substituting the expressions from steps 4 and 5, we get:
   \[
   g(k^2) = g(k) + g(k)
   \]
   This simplifies to:
   \[
   g(k^2) = 2g(k)
   \]

7. Let \( k = e^t \). Then \( g(e^{2t}) = 2g(e^t) \). Define \( h(t) = g(e^t) \). Then:
   \[
   h(2t) = 2h(t)
   \]

8. This functional equation \( h(2t) = 2h(t) \) suggests that \( h(t) \) is a linear function. Assume \( h(t) = ct \) for some constant \( c \). Then:
   \[
   g(e^t) = ct
   \]
   Substituting back \( t = \ln k \), we get:
   \[
   g(k) = c \ln k
   \]

9. Therefore:
   \[
   I(a, b) = g\left(\frac{b}{a}\right) = c \ln \left(\frac{b}{a}\right)
   \]

10. Now, consider the integral \( \int_a^b f(x) \, dx \). We need to find \( f(x) \) such that:
    \[
    \int_a^b f(x) \, dx = c \ln \left(\frac{b}{a}\right)
    \]

11. Differentiate both sides with respect to \( b \):
    \[
    f(b) = \frac{d}{db} \left( c \ln \left(\frac{b}{a}\right) \right) = \frac{c}{b}
    \]

12. Therefore, we have:
    \[
    f(x) = \frac{c}{x}
    \]

\(\blacksquare\)

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