Showing that $f(x) = \dfrac{x}{(2\ln x)^2}$ is an increasing function for $x \ge 8$ I apologize for the repetition.  I asked a similar question here before.
I was trying to generalize the result.  Does the following reasoning also work to show that $f(x) = \dfrac{x}{(2\ln x)^2}$ is an increasing function for $x \ge 8$
Please let me know if any of these steps are wrong:
(1)  Using the quotient rule with $g(x) = x$ and $h(x) = (2\ln x)^2$:
$$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$$
(2) Using the exponent rule for derivatives with $s(x) = 2\ln x$:
$$h'(x) = (s(x)^2)' = s(x)^2\left(s'(x)\dfrac{2}{s(x)}\right) = 2s(x)s'(x)$$
(3) $s'(x) = \dfrac{2}{x}$ so that:
$$h'(x) = \dfrac{8\ln(x)}{x}$$
(4) With $g'(x) = 1$, it follows that:
$$f'(x) = \dfrac{(2\ln x)^2 - \frac{8x\ln(x)}{x}}{(2\ln x)^4} = \dfrac{(2\ln x) - 4}{(2\ln x)^3}$$
(5)  It is increasing at $x\ge 8$ since:
$$\dfrac{(2\ln 8) - 4}{(2\ln(8))^3} > 0.0022 > 0$$
Are these steps correct? 

Edit:  I changed step(5) to $x\ge 8$ since that is my goal.
It looks like my result may be correct for $x=8$ but insufficient for $x \ge 8$.

Edit 2:  Made a fix based on John Omielan's comment.
 A: You have almost reached there. You will just have to modify your step 5.
$$f'(x)=\frac{2(ln\ x-2)}{8(ln\ x)^3}$$
$$f'(x)=\frac{ln\ x-2}{4(ln\ x)^3}$$ 
We know that $ln\ x>0\  \forall x>1$ and hence for all $x\ge8$. Therefore its cube, $(ln\ x)^3$ is also positive $\forall x\ge8$. That makes the denominator of $f'(x)$ positive $\forall x\ge8$.
Also, $ln\ x$ is a strictly increasing function as its first derivative, $\frac{1}{x}$ is positive in all its domain, i.e., $\forall x>0$.
We also know that $ln\ 8=2.079... >2$, and by the definition of an increasing function we can conclude that $ln\ x\ge ln\ 8>2, \forall x\ge 8$.
$$\implies ln\ x-2>0, \forall x\ge8$$
Hence, we also established that the numerator of $f'(x)$ is positive $\forall x\ge8$.
A rational function is positive if and only if its numerator and denominator functions have the same sign.
Since both its numerator and its denominator are always positive, and hence have same sign, $\forall x\ge8$ we can conclude that $f'(x)$ is positive in the same interval.
Hence proved that $f(x)$ is increasing $\forall x\ge8$
