Find all differentiable functions $f, g:[0,\infty) \to \mathbb{R}$ and the real constant $k\geq 0$ such that 
\begin{align*} f(x) &=k+ \int_0^x \frac{g(t)}{f(t)}dt \\ 
g(x) &= -k-\int_0^x f(t)g(t) dt \end{align*} 
and $f(0)=k, f'(0)=-k^2/3$ and also $f(x)\neq 0$ for all $x\geq 0$.\\ \\ 
[i] (Nora Gavrea)[/i]
**
    Checking the initial conditions:
    \[
    f(0) = \frac{3k}{3} = k
    \]
    \[
    g(0) = -\frac{9k^3}{3^3} = -\frac{k^3}{3}
    \]
    Given \( g(0) = -k \):
    \[
    -\frac{k^3}{3} = -k \implies k^3 = 3k \implies k = \sqrt{3}
    \]

The final answer is \( \boxed{ f(x) = \frac{3\sqrt{3}}{3 + \sqrt{3}x} } \) and \( g(x) = -\frac{27}{(3 + \sqrt{3}x)^3} \).