
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
First, let us consider the function \(\varphi(t)=t + \frac{1}{t}\). The function \(\varphi(t)\) is defined for all \(t \neq 0\). We will find the extrema of the function \(\varphi(t)\). To do this, we will find the intervals where the derivative of the function \(\varphi(t)\) is constant: \(\varphi^{\prime}(t)=1-\frac{1}{t^{2}}=\frac{(t-1)(t+1)}{t^{2}}, \quad \varphi^{\prime}(t)=0 \quad\) at \(t= \pm 1\). When passing through the point \(t=-1\), the derivative \(\varphi^{\prime}(t)\) changes sign from plus to minus, which means that \(t=-1\) is a maximum point, \(\varphi_{\max }=\varphi(-1)=-2\). When passing through the point \(t=1\), the derivative \(\varphi^{\prime}(t)\) changes sign from minus to plus, which means that \(t=1\) is a minimum point, \(\varphi_{\min }=\varphi(1)=2\). The range of this function is \((-\infty ;-2] \cup[2 ;+\infty)\).

Next, consider the function \(g(x)=x^{3}+\frac{1}{x^{3}}=\varphi(x^{3})\). Since the function \(t=x^{3}\) is increasing on the interval \((-\infty ;+\infty)\) and takes all real values, the range of the function \(\varphi(x^{3})\), and thus \(g(x)\), is \((-\infty ;-2] \cup[2 ;+\infty)\), with \(g_{\max }=g(-1)=-2\) and \(g_{\min }=g(1)=2\).

For the same reason, the range of the function \(g(\ln x)\) is also \((-\infty ;-2] \cup[2 ;+\infty)\). The range of the function \(g(g(\ln x))\) is \((-\infty ;-\frac{65}{8}] \cup[\frac{65}{8} ;+\infty)\), and for the function \(\frac{16 g(g(\ln x))}{65}\), the range is \((-\infty ;-2] \cup[2 ;+\infty)\). Thus, \(g\left(\frac{16 g(g(\ln x))}{65}\right) \in (-\infty ;-\frac{65}{8}] \cup[\frac{65}{8} ;+\infty)\).

Therefore, the range \(E_f\) of the function \(f(x)=\frac{1}{g\left(\frac{16 g(g(\ln x))}{65}\right)}\) is \([-\frac{8}{65} ; 0) \cup (0 ; \frac{8}{65}]\).

Present the answer in LaTex format: \boxed{Your answer}
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