$y=\frac{(2x+3)^9}{\sqrt x(x^2-x)^6}$
I switched it to $\ln(y)=9\ln(2x+3)-6x^{1/2}\ln(x^2-x)$ and then used the log rules for derivatives I know and the product rule on the right side and wound up with $$y'=y\left[\frac{18}{2x+3}-\frac{3\ln(x^2-x)}{x^{1/2}}+\frac{6x^{1/2}(2x-1)}{x^2-x}\right]$$ The $\frac{18}{2x+3}$ is the correct form but the rest is way off from the form my answer is supposed to be in. I'm not quite sure what else to do with it though...
To solve for the derivative of $y=\frac{(2x+3)^9}{\sqrt x(x^2-x)^6}$, we'll be using logarithmic differentiation which, as you said, you were having trouble with. In taking the natural logarithm on both sides, we now have
$$\ln(y)=9\ln(2x+3)-\ln(x^{1/2}(x^2-x)^6).$$
We know the product logarithm rule, $\ln(ab)=\ln(a)+\ln(b)$, so we'll use that on our second term in the RHS:
$$\ln(y)=9\ln(2x+3)-(\ln(x^{1/2})+\ln((x^2-x)^6).$$
We'll simplify down some of our RHS terms now:
$$\ln(y)=9\ln(2x+3)-1/2\ln(x)-6\ln(x^2-x).$$
$-6\ln(x^2-x)$ can be simplified to $-6\ln(x(x-1)),$ or $-6\ln(x)-6\ln(x-1).$ We can now obtain
$$\ln(y)=9\ln(2x+3)-1/2\ln(x)-6\ln(x)-6\ln(x-1).$$
And I'll simplify $-1/2\ln(x)-6\ln(x) = -13/2\ln(x)$.
$$\ln(y)=9\ln(2x+3)-13/2\ln(x)-6\ln(x-1).$$
Using logarithmic differentials, we can do this:
$$y'=y\left(\frac{18}{2x+3}-\frac{13/2}{x}-\frac{6}{x-1}\right)$$
Substitute for $y$, giving
$$y'=\frac{(2x+3)^9}{\sqrt x(x^2-x)^6}\left(\frac{18}{2x+3}-\frac{13/2}{x}-\frac{6}{x-1}\right).$$
I'll let you do the rest.