Find continuous functions $ f(x),\ g(x)$ which takes positive value for any real number $ x$, satisfying  $ g(x)\equal{}\int_0^x f(t)\ dt$ and $ \{f(x)\}^2\minus{}\{g(x)\}^2\equal{}1$.
1. Given the problem, we start with the equations:
   \[
   g(x) = \int_0^x f(t) \, dt
   \]
   and
   \[
   \{f(x)\}^2 - \{g(x)\}^2 = 1.
   \]

2. First, differentiate \( g(x) \) with respect to \( x \):
   \[
   g'(x) = f(x).
   \]

3. Substitute \( g'(x) \) into the second equation:
   \[
   \{g'(x)\}^2 - \{g(x)\}^2 = 1.
   \]

4. This simplifies to:
   \[
   \left(g'(x)\right)^2 + \left(g(x)\right)^2 = 1.
   \]

5. Differentiate both sides of the equation with respect to \( x \):
   \[
   2 g'(x) g''(x) + 2 g(x) g'(x) = 0.
   \]

6. Factor out \( 2 g'(x) \):
   \[
   g'(x) \left(g''(x) + g(x)\right) = 0.
   \]

7. Since \( g'(x) \neq 0 \) (as \( g'(0) = \pm 1 \)), we have:
   \[
   g''(x) + g(x) = 0.
   \]

8. The general solution to the differential equation \( g''(x) + g(x) = 0 \) is:
   \[
   g(x) = A \cosh(x) + B \sinh(x).
   \]

9. Given \( g(0) = 0 \):
   \[
   g(0) = A \cosh(0) + B \sinh(0) = A \cdot 1 + B \cdot 0 = A.
   \]
   Therefore, \( A = 0 \).

10. Thus, the solution simplifies to:
    \[
    g(x) = B \sinh(x).
    \]

11. Given \( g'(0) = \pm 1 \):
    \[
    g'(x) = B \cosh(x).
    \]
    At \( x = 0 \):
    \[
    g'(0) = B \cosh(0) = B \cdot 1 = B.
    \]
    Therefore, \( B = \pm 1 \).

12. Hence, the function \( g(x) \) is:
    \[
    g(x) = \pm \sinh(x).
    \]

13. Since \( g'(x) = f(x) \), we have:
    \[
    f(x) = \pm \cosh(x).
    \]

The final answer is \( \boxed{ f(x) = \pm \cosh(x) } \) and \( g(x) = \pm \sinh(x) \).