I hope that the following comments settle some of your questions. Unfortunately, it is not clear what notation you use for the Chain Rule.
1) Suppose that $H(x)=\ln(5x)$. We want to find H'(x). Note that $\ln(5x)$ is "function of a function of $x$." Specifically, $\ln(5x)=f(g(x)$, where $g(x)=5x$ and $f(u)=\ln(u)$.
Therefore, by the Chain Rule, we have H'(x)=g'(x)f'(g(x)). Note that g'(x)=5 and f'(u)=\dfrac{1}{u}. So f'(g(x))=\dfrac{1}{5x}.
Now put things together. We have H'(x)=(5)\dfrac{1}{5x}. This simplifies to $\dfrac{1}{x}$.
In this case, we could have avoided the Chain Rule, by noting that $H(x)=\ln(5x)=\ln(5)+\ln(x)$. Differentiate, using the fact that $\ln(5)$ is a constant and therefore its derivative is $0$. We get H'(x)=\dfrac{1}{x}.
2) Let $H(x)=\ln(3x^2-1)$. We want H'(x). Note that $H(x)=f(g(x))$ where $g(x)=3x^2-1$ and $f(u)=\ln(u)$. Use the Chain Rule like before. We have g'(x)=6x and f'(u)=\dfrac{1}{u}. Putting things together like before, we get H'(x)=\dfrac{6x}{3x^2-1}. No useful simplification is available.
Important note: It looks as if your post asks why we can't treat $\ln(3x^2-1)$ as $\dfrac{\ln(3x^2)}{\ln(1)}$.
The reason that we can't is that $\ln(3x^2-1)$ is not equal to $\dfrac{\ln(3x^2)}{\ln(1)}$. The logarithm of a product is a sum of logarithms. the logarithm of a quotient is a difference of logarithms. So for example, $\ln((x^2+1)(x^2+5))=\ln(x^2+1)+\ln(x^2+5)$, and $\ln(7/5)=\ln(7)-\ln(5)$.
But the logarithm of a sum can't be "simplified" in a pleasant way, and neither can the logarithm of a difference. You can check for yourself, with a calculator, that $\ln(7-2)$ is not equal to $\dfrac{\ln(7)}{\ln(2)}.$
If you use the FALSE "rules" $\ln(a+b)=\ln(a)+\ln(b)$ or $\ln(a-b)=\dfrac{\ln(a)}{\ln(b)}$, it will cause you grief over and over.