You could apply a clever Algebra trick to avoid using the quotient rule,
[tex]\rm y=\dfrac{log(x)-1+1}{log(x)-1}=\dfrac{log(x)-1}{log(x)-1}+\dfrac{1}{log(x)-1}[/tex]
[tex]\rm y=1+\dfrac{1}{log(x)-1}[/tex]
and apply power rule into chain rule from that point.
But this problem was likely designed to teach you about quotient rule so let's do it that way.
Let's start by "setting up" our quotient rule:
[tex]\rm y'=\dfrac{(log x)'(log x-1)-log x(log x-1)'}{(log x-1)^2}[/tex]
If this log notation is not intended to be natural log, then we'll have a little bit of extra work. Our change of base identity allows us to rewrite log base 10 in terms of the natural log,
[tex]\rm log(x)=\dfrac{ln x}{ln10}[/tex]
so let's apply this to our problem,
[tex]\rm y'=\dfrac{\left(\frac{ln x}{ln 10}\right)'(log x-1)-log x\left(\dfrac{ln x}{ln 10}-1\right)'}{(log x-1)^2}[/tex]
Derivative of ln(x) gives us 1/x in each case,
[tex]\rm y'=\dfrac{\left(\frac{1}{x ln 10}\right)(log x-1)-log x\left(\dfrac{1}{x ln 10}\right)}{(log x-1)^2}[/tex]
Factor the 1/(x ln10) out of each term in the numerator,
[tex]\rm y'=\left(\dfrac{1}{x ln 10}\right)\dfrac{log x-1-log x}{(log x-1)^2}[/tex]
and combine like-terms,
[tex]\rm y'=\dfrac{-1}{x(log x-1)^2ln 10}[/tex]
Lemme know if you're confused with any of the steps.
