I have the function:
$\frac{x-1}{x^3}$
I differentiated this using the quotient rule and got:
$-\frac{2x-3}{x^4}$
What I need to do and can't seem to work out is how to differentiate the same function by (a) chain rule and then (b) "applying the properties of logarithms to break up the expression into simpler logarithms first".
Let $$y=\frac{x-1}{x^3}$$
Then $$\log y=\log (x-1)- \log (x^3)$$
$$\Rightarrow \log y=\log (x-1)- 3\log (x)$$
$$\Rightarrow \frac 1y\frac{dy}{dx}=\frac 1{x-1}-\frac 3x$$
$$\Rightarrow \frac{x^3}{x-1}\frac{dy}{dx}=\frac 1{x-1}-\frac 3x$$
$$\Rightarrow \frac{dy}{dx}=\frac 1{x^3}-\frac {3(x-1)}{x^4}$$
$$\Rightarrow \frac{dy}{dx}=\frac x{x^4}-\frac {3x-3}{x^4}$$
$$\Rightarrow \frac{dy}{dx}=-\frac {2x-3}{x^4}$$ which is the same as you have found.
By chain rule I expect they mean for you to recognise that $$y=\frac{x-1}{x^3}=\frac{1}{x^2}-\frac{1}{x^3}=(x)^{-2}-(x)^{-3}$$
$$\Rightarrow \frac{dy}{dx}=(-2)(x)^{-3}-(-3)(x)^{-4}$$
$$\Rightarrow \frac{dy}{dx}=-\frac {2}{x^3}+\frac {3}{x^4}=-\frac {2x-3}{x^4}$$ which is the same again.