Let the domain of the function $$f(x)=\log_{3}\log_{5}\log_{7}(9x-x^{2}-13)$$ be the interval (m, n). Let the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ have eccentricity $$\frac{n}{3}$$ and the length of the latus rectum $$\frac{8m}{3}$$. Then $$b^{2}-a^{2}$$ is equal to:

$$9x - x^{2} - 13 \gt 0$$
$$\log_{7}(9x - x^{2} - 13) \gt 1 \;\;\Longrightarrow\;\; 9x - x^{2} - 13 \gt 7$$

The second inequality is stronger, so we keep it:
$$9x - x^{2} - 13 \gt 7 \;\;\Longrightarrow\;\; 9x - x^{2} - 20 \gt 0$$
$$-x^{2} + 9x - 20 \gt 0 \;\;\Longrightarrow\;\; x^{2} - 9x + 20 \lt 0$$
$$x^{2} - 9x + 20 = (x-4)(x-5)$$

 the domain is the open interval$$(4,\,5)$$
Thus $$m = 4$$ and $$n = 5$$.

For the hyperbola $$\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$$:

• Eccentricity $$e = \dfrac{n}{3} = \dfrac{5}{3}$$
• Length of the latus rectum $$L = \dfrac{8m}{3} = \dfrac{8 \times 4}{3} = \dfrac{32}{3}$$

The standard results for such a hyperbola are

$$e^{2} = 1 + \dfrac{b^{2}}{a^{2}}$$
$$L = \dfrac{2b^{2}}{a}$$

$$\left(\dfrac{5}{3}\right)^{2} = 1 + \dfrac{b^{2}}{a^{2}} \;\;\Longrightarrow\;\; \dfrac{25}{9} = 1 + \dfrac{b^{2}}{a^{2}}$$

$$\dfrac{b^{2}}{a^{2}} = \dfrac{25}{9} - 1 = \dfrac{16}{9} \;\;\Longrightarrow\;\; b^{2} = \dfrac{16}{9}\,a^{2} \;$$

Using the latus-rectum length:

$$\dfrac{2b^{2}}{a} = \dfrac{32}{3} \;\;\Longrightarrow\;\; b^{2} = \dfrac{16}{3}\,a \;$$

$$\dfrac{16}{9}\,a^{2} = \dfrac{16}{3}\,a \;\;\Longrightarrow\;\; \dfrac{1}{9}a^{2} = \dfrac{1}{3}a \;\;\Longrightarrow\;\; a^{2} = 3a \;\;\Longrightarrow\;\; a(a-3)=0$$

Since $$a\gt0$$, we take $$a = 3$$, giving $$a^{2} = 9$$.

Substitute back to find $$b^{2}$$:

$$b^{2} = \dfrac{16}{9}\,a^{2} = \dfrac{16}{9}\times9 = 16$$

$$b^{2} - a^{2} = 16 - 9 = 7$$