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Let the equation of the ellipse be:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (b > a)$

Given that the coordinates of the foci are $(0, \pm 3)$ and the equations of the directrices are $3y = \pm 25$, i.e., $y = \pm \frac{25}{3}$.

Comparing with the standard forms for an ellipse with major axis along the y-axis:
$be = 3$
$\frac{b}{e} = \frac{25}{3}$

Multiplying these two equations:
$(be) \times \left(\frac{b}{e}\right) = 3 \times \frac{25}{3} \implies b^2 = 25$

Thus, $e^2 = \frac{(be)^2}{b^2} = \frac{3^2}{25} = \frac{9}{25}$.

Now, for $b > a$:
$a^2 = b^2(1 - e^2) = 25\left(1 - \frac{9}{25}\right) = 25 \times \frac{16}{25} = 16$

Therefore, the required equation of the ellipse is:
$\frac{x^2}{16} + \frac{y^2}{25} = 1 \quad \text{or} \quad 25x^2 + 16y^2 = 400$