Let $x$ be a number such that $x +\frac{1}{x}=-1$. Determine the value of  $x^{1994} +\frac{1}{x^{1994}}$.
1. Let \( x \) be a number such that \( x + \frac{1}{x} = -1 \). We need to determine the value of \( x^{1994} + \frac{1}{x^{1994}} \).

2. Define \( a_n = x^n + \frac{1}{x^n} \). We aim to find a recurrence relation for \( a_n \).

3. Using the given equation \( x + \frac{1}{x} = -1 \), we can derive the recurrence relation:
   \[
   a_{n+2} = a_{n+1}a_1 - a_n
   \]
   where \( a_1 = x + \frac{1}{x} = -1 \).

4. Calculate the initial terms:
   \[
   a_1 = -1
   \]
   \[
   a_2 = x^2 + \frac{1}{x^2} = \left( x + \frac{1}{x} \right)^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1
   \]

5. Using the recurrence relation, calculate the next terms:
   \[
   a_3 = a_2 a_1 - a_1 = (-1)(-1) - (-1) = 1 + 1 = 2
   \]
   \[
   a_4 = a_3 a_1 - a_2 = 2(-1) - (-1) = -2 + 1 = -1
   \]
   \[
   a_5 = a_4 a_1 - a_3 = (-1)(-1) - 2 = 1 - 2 = -1
   \]

6. Observe that the sequence \( a_n \) repeats every 3 terms starting from \( a_2 \):
   \[
   a_2 = -1, \quad a_3 = 2, \quad a_4 = -1, \quad a_5 = -1, \quad a_6 = 2, \quad \ldots
   \]

7. Since the sequence repeats every 3 terms, we find the position of \( a_{1994} \) in the cycle:
   \[
   1994 \mod 3 = 2
   \]
   Therefore, \( a_{1994} = a_2 = -1 \).

8. Alternatively, we can solve the given equation \( x + \frac{1}{x} = -1 \) by multiplying both sides by \( x \):
   \[
   x^2 + 1 = -x \implies x^2 + x + 1 = 0
   \]
   The roots of this equation are the non-real cube roots of unity, \( x = e^{2\pi i / 3} \) or \( x = e^{-2\pi i / 3} \).

9. Since \( x \) is a cube root of unity, we have:
   \[
   x^3 = 1 \implies x^{1994} = x^{3 \cdot 664 + 2} = (x^3)^{664} x^2 = 1^{664} x^2 = x^2
   \]
   Similarly,
   \[
   \frac{1}{x^{1994}} = \frac{1}{x^2}
   \]

10. Therefore,
    \[
    x^{1994} + \frac{1}{x^{1994}} = x^2 + \frac{1}{x^2} = a_2 = -1
    \]

Conclusion:
\[
\boxed{-1}
\]