To find the eccentricity of the ellipse, we need to first put the given equation in the standard form of an ellipse, which is:

(x^2/a^2) + (y^2/b^2) = 1

where 'a' and 'b' are the semi-major and semi-minor axes of the ellipse, respectively.

1. Rearrange the equation:
9x^2 + 5y^2 + 30y = 0

2. Complete the square for the y terms:
9x^2 + 5(y^2 + 6y) = 0

3. Divide the coefficient of x^2 by the constant term to determine 'a':
a^2 = -30/9
a^2 = -10/3
a = sqrt(-10/3) (imaginary)

4. Divide the coefficient of y^2 by the constant term to determine 'b':
b^2 = -5/9
b = sqrt(-5/9) (imaginary)

Since 'a' and 'b' are imaginary, this means that the given equation does not represent a real ellipse, but rather a degenerate ellipse or a pair of intersecting lines.

5. Calculate the eccentricity (e):
The eccentricity of an ellipse is defined as the square root of (1 - (b^2/a^2)).

e = sqrt(1 - (b^2/a^2))
e = sqrt(1 - (-5/9) / (-10/3))
e = sqrt(1 + 5/6)
e = sqrt(11/6)

Therefore, the eccentricity of the given ellipse is sqrt(11/6), which is approximately 0.8819. None of the given options (a, b, c, d) match this value, so the correct answer is None of these.