Given equation:\[4x^2 - 9y^2 - 1 = 0\]Rearranging the equation yields:\[\frac{x^2}{\frac{1}{4}} - \frac{y^2}{\frac{1}{9}} = 1\]\[\frac{x^2}{\left(\frac{1}{2}\right)^2} - \frac{y^2}{\left(\frac{1}{3}\right)^2} = 1\]Comparing with the standard hyperbola form:\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]we identify:\[a = \frac{1}{2}, \quad b = \frac{1}{3}\]The eccentricity \(e\) of a hyperbola is calculated as:\[e = \sqrt{1 + \frac{b^2}{a^2}}\]Substituting the values of \(a\) and \(b\):\[= \sqrt{1 + \frac{\frac{1}{9}}{\frac{1}{4}}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}\]The foci are located at:\[(\pm ae, 0) = \left( \pm \frac{1}{2} \times \frac{\sqrt{13}}{3}, 0 \right)\]\[= \left( \pm \frac{\sqrt{13}}{6}, 0 \right)\]Therefore, the foci of the hyperbola are:\[\left( \pm \frac{\sqrt{13}}{6}, 0 \right)\]