If there a function $f(x)$ such that following three conditions can be satisfied? $$\bigg(\frac{f\big(x^{c(x)}\big)}{\log f\big(x^{c(x)}\big)}\bigg)^{\log f(x^{c(x)})}= c(x)(\log x) x^{c(x)}$$ $$c(x)=\frac{\log f(x^{c(x)})}{2 f(x^{c(x)})-\log f(x^{c(x)})}$$ $$1\leq\log f\big(x^{c(x)}\big)<f\big(x^{c(x)}\big)\leq\frac{({{1+c(x)}})\log x}{4\log(1+c(x))-4\log2+4\log\log x}$$
where $c(x)$ is given by: $$0<c(x)\leq\frac{\log\log x}{(\log x)^\beta}$$ at some $\beta\in(0,1)$.
Partial Solution - I'm almost done, but it looks like you're getting antsy, so here's what I have so far.
Look at the first condition. Note that this is entirely a function of $x^{c(x)}$.
Let $x^{c(x)} = g(x)$. The condition then becomes
$$\left(\frac{f(g(x))}{\log f(g(x))}\right)^{\log f(g(x))} = g(x)\log(g(x))$$
Letting $u=g(x)$:
$$\left(\frac{f(u)}{\log f(u)}\right)^{\log f(u)} = u\log u$$
Taking $\log$ of each side:
$$(\log f(u))^2-(\log f(u))(\log\log f(u)) = \log u - \log\log u$$
Define
$$h(a) = a^2-a\log a$$
So, we then get that, for all $u$ in the image of $x^{c(x)}$,
$f(u) = \exp(h^{-1}(\log u-\log\log u))$
Since we only care about the value of $f$ at $x^{c(x)}$, we can assume that
$$f\left(x^{c(x)}\right) = \exp\bigg(h^{-1}\big((c(x)\log x) - (\log c(x)) - (\log\log x)\big)\bigg)$$
We now have our function $f\left(x^{c(x)}\right)$, and all that remains is to see whether it satisfies the other properties. We have
$$c(x) = \frac{\log f(u)}{2f(u)-\log f(u)}$$
using the same $u$ as before. Solving for $f(u)$ in terms of $c(x)$:
$$2f(u)c(x) - c(x)(\log f(u)) = \log f(u)$$
$$\frac{2c(x)}{c(x)+1}f(u) = \log f(u)$$
Letting $a = \frac{2c(x)}{c(x)+1}$:
$$af(u) = \log f(u)$$
$$e^{af(u)} = f(u)$$
$$-ae^{af(u)} = -af(u)$$
$$-a = (-af(u))e^{(-af(u))}$$
Using the Lambert W function defined as the inverse of $h(x) = xe^x$:
$$-af(u) = W(-a)$$
$$f(u) = \frac{W(-a)}{-a}$$
$$f\left(x^{c(x)}\right) = -\frac{W\left(-\frac{2c(x)}{c(x)+1}\right)}{\frac{2c(x)}{c(x)+1}}$$
But we have from before that
$$f\left(x^{c(x)}\right) = \exp\bigg(h^{-1}\big((c(x)\log x) - (\log c(x)) - (\log\log x)\big)\bigg)$$
so, for a solution to exist, we must have that
$$-\frac{(c(x)+1)W\left(-\frac{2c(x)}{c(x)+1}\right)}{2c(x)} = \exp\bigg(h^{-1}\big((c(x)\log x) - (\log c(x)) - (\log\log x)\big)\bigg)$$
$$\log(c(x)+1)-\log(c(x))-\log(2)+\log\left(-W\left(-\frac{2c(x)}{c(x)+1}\right)\right) = h^{-1}\big((c(x)\log x) - (\log c(x)) - (\log\log x)\big)$$
$$h\left(\log(c(x)+1)-\log(c(x))-\log(2)+\log\left(-W\left(-\frac{2c(x)}{c(x)+1}\right)\right)\right) = (c(x)\log x) - (\log c(x)) - (\log\log x)$$
By the definition of $h$:
$$\left(\log(c(x)+1)-\log(c(x))-\log(2)+\log\left(-W\left(-\frac{2c(x)}{c(x)+1}\right)\right)\right)^2-\left(\log(c(x)+1)-\log(c(x))-\log(2)+\log\left(-W\left(-\frac{2c(x)}{c(x)+1}\right)\right)\right)\log\left(\log(c(x)+1)-\log(c(x))-\log(2)+\log\left(-W\left(-\frac{2c(x)}{c(x)+1}\right)\right)\right)= (c(x)\log x) - (\log c(x)) - (\log\log x)$$