The function $ f(x,y,z)\equal{}\frac{x^2\plus{}y^2\plus{}z^2}{x\plus{}y\plus{}z}$ is defined for every $ x,y,z \in R$ whose sum is not 0. Find a point $ (x_0,y_0,z_0)$ such that $ 0 < x_0^2\plus{}y_0^2\plus{}z_0^2 < \frac{1}{1999}$ and $ 1.999 < f(x_0,y_0,z_0) < 2$.
1. **Define the function and constraints:**
   The function is given by:
   \[
   f(x, y, z) = \frac{x^2 + y^2 + z^2}{x + y + z}
   \]
   We need to find a point \((x_0, y_0, z_0)\) such that:
   \[
   0 < x_0^2 + y_0^2 + z_0^2 < \frac{1}{1999}
   \]
   and
   \[
   1.999 < f(x_0, y_0, z_0) < 2
   \]

2. **Introduce small perturbations:**
   Let \(\delta \approx 0^+\) and \(\epsilon \approx 0^+\). Define:
   \[
   \kappa = \frac{2 - 4\delta - \sqrt{(4\delta - 2 + \epsilon)^2 - 4 \cdot 6\delta^2}}{2}
   \]

3. **Claim the point \((x_0, y_0, z_0)\):**
   We claim that:
   \[
   (x_0, y_0, z_0) = (2\delta + \kappa, -\delta, -\delta)
   \]
   satisfies the given conditions.

4. **Approximate \(\kappa\):**
   Since \(\delta \approx 0^+\) and \(\epsilon \approx 0^+\), we have:
   \[
   \kappa = \frac{2 - 4\delta - \sqrt{(4\delta - 2 + \epsilon)^2 - 4 \cdot 6\delta^2}}{2}
   \]
   Approximating further:
   \[
   \kappa \approx \frac{2 - 4 \cdot 0 - \sqrt{(4 \cdot 0 - 2 + 0)^2 - 4 \cdot 6 \cdot 0^2}}{2} = \frac{2 - \sqrt{2^2}}{2} = 0
   \]
   Thus, \(\kappa \approx 0\).

5. **Evaluate \(x_0, y_0, z_0\):**
   Therefore:
   \[
   x_0 = 2\delta + \kappa \approx 0, \quad y_0 = z_0 = -\delta \approx 0^-
   \]
   Hence:
   \[
   x_0^2, y_0^2, z_0^2 \approx 0^+
   \]

6. **Verify the quadratic equation:**
   \(\kappa\) is a root of the quadratic equation:
   \[
   x^2 + x(4\delta - 2 + \epsilon) + 6\delta^2 = 0
   \]
   Thus, \(\kappa\) satisfies:
   \[
   \kappa^2 + \kappa(4\delta - 2 + \epsilon) + 6\delta^2 = 0
   \]

7. **Evaluate the function \(f(x_0, y_0, z_0)\):**
   This is equivalent to:
   \[
   \frac{(2\delta + \kappa)^2 + 2\delta^2}{\kappa} = 2 - \epsilon
   \]
   But the left-hand side (LHS) is precisely \(f(x_0, y_0, z_0)\).

8. **Conclusion:**
   Hence, our claim is true, and the point \((x_0, y_0, z_0)\) satisfies the given conditions.

The final answer is \( \boxed{ (x_0, y_0, z_0) = (2\delta + \kappa, -\delta, -\delta) } \)