First, the function $f(x) = \sin^{-1}(\log_{3x}(\frac{6 + 2 \log_3{x}}{-5x}))$ has several restrictions :

Since the arcsine function $\sin^{-1}(x)$ is only defined for $-1\leq x\leq 1$, this means that $\log_{3x}(\frac{6 + 2 \log _3 x}{-5 x})$ must be between -1 and 1.
For the logarithm to be defined, $3x > 0 \Rightarrow x > 0$ .......(1)

because the base of a logarithm must be greater than 0 and not equal to 1. Also, $x \neq \frac{1}{3}$ .......(2)

as the base cannot be 1.
Moreover, the inner function of the logarithm $\frac{6 + 2 \log_3{x}}{-5x}$ must be greater than 0.

$$ \Rightarrow $$ $6+2 \log _3 x<0 \quad(\because x>0)$

$$ \begin{aligned} \Rightarrow & \log _3 x<-3 \\\\ \Rightarrow & x<3^{-3} \\\\ \Rightarrow & x<\frac{1}{27} .........(3) \end{aligned} $$

$$ \Rightarrow \text { From (1), (2), (3), } 0 < x < \frac{1}{27} $$

The next step is to solve the inequality for $-1 \leq \log_{3x}(\frac{6+2 \log _3 x}{-5 x}) \leq 1$. To do this, we make the observation that for the logarithmic part to be within $[-1,1]$, it must be true that $3x \leq \frac{6+2 \log_3 x}{-5x} \leq \frac{1}{3x}$.

Solving the inequality $15x^2 + 6 + 2 \log_3 x \geq 0$, we find that $x \in(0, \frac{1}{27})$ ........(4)

Likewise, solving the inequality $6+2 \log_3 x + \frac{5}{3} \geq 0$, we find that $x \geq 3^{-\frac{23}{6}}$ ........(5)

Combining Equations (3), (4), and (5), the intersection of all these intervals is $x \in[3^{-\frac{23}{6}}, \frac{1}{27})$.

Now, consider the function $g(x) = x - [x]$, where $[x]$ is the greatest integer function. For this function, the range is the fractional part of $x$. In this case, the range $(\alpha, \beta)$ is given by the minimum and maximum possible values of $x$ in its domain. Hence, $\alpha = 3^{-\frac{23}{6}}$ and $\beta = \frac{1}{27}$.

Finally, substitute these values into the equation $\alpha^{2}+\frac{5}{\beta}$:

$\alpha^{2}+\frac{5}{\beta} = (3^{-\frac{23}{6}})^2 + \frac{5}{\frac{1}{27}} = 3^{-\frac{23}{3}} + 135$ = 135.0002198.

Since $3^{-\frac{23}{3}}$ = 0.0002198 is an extremely small number, it's approximately 0. So, $\alpha^{2}+\frac{5}{\beta} \approx 135$.

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