If $\alpha$ and $\beta$ are the roots of the equation $3x^2+x-1=0$, where $\alpha>\beta$, find the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$.

$ \textbf{(A) }\frac{7}{9}\qquad\textbf{(B) }-\frac{7}{9}\qquad\textbf{(C) }\frac{7}{3}\qquad\textbf{(D) }-\frac{7}{3}\qquad\textbf{(E) }-\frac{1}{9} $
1. Given the quadratic equation \(3x^2 + x - 1 = 0\), we need to find the roots \(\alpha\) and \(\beta\).
2. By Vieta's formulas, for a quadratic equation \(ax^2 + bx + c = 0\), the sum of the roots \(\alpha + \beta\) is given by \(-\frac{b}{a}\) and the product of the roots \(\alpha \beta\) is given by \(\frac{c}{a}\).
3. For the given equation \(3x^2 + x - 1 = 0\):
   - \(a = 3\)
   - \(b = 1\)
   - \(c = -1\)
4. Using Vieta's formulas:
   \[
   \alpha + \beta = -\frac{b}{a} = -\frac{1}{3}
   \]
   \[
   \alpha \beta = \frac{c}{a} = \frac{-1}{3}
   \]
5. We need to find the value of \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\).
6. Notice that:
   \[
   \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta}
   \]
7. To find \(\alpha^2 + \beta^2\), we use the identity:
   \[
   \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta
   \]
8. Substituting the values from Vieta's formulas:
   \[
   (\alpha + \beta)^2 = \left(-\frac{1}{3}\right)^2 = \frac{1}{9}
   \]
   \[
   2\alpha \beta = 2 \left(-\frac{1}{3}\right) = -\frac{2}{3}
   \]
9. Therefore:
   \[
   \alpha^2 + \beta^2 = \frac{1}{9} - \left(-\frac{2}{3}\right) = \frac{1}{9} + \frac{2}{3} = \frac{1}{9} + \frac{6}{9} = \frac{7}{9}
   \]
10. Now, substituting back into the expression for \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\):
    \[
    \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{\frac{7}{9}}{-\frac{1}{3}} = \frac{7}{9} \times -3 = -\frac{21}{9} = -\frac{7}{3}
    \]

The final answer is \(\boxed{-\frac{7}{3}}\)