Using proportion
Still on moving escalator, travel 1 escalator in 60 seconds = $\frac1{60}$ escalator per second
Walking on still escalator, travel 1 escalator in 90 seconds = $\frac1{90}$ escalator per second
Walking on moving escalator, travel $\frac1{60}+\frac1{90}$ escalator per second
$=\frac{3}{180}+\frac2{180}$
$=\frac5{180}$
$=\frac1{36}$ escalator per second = 1 escalator in 36 seconds.
Using a distance-time graph
Length of escalator is $d$.
Walking up the moving escalator: add the distances.
If Aimee walks for $60$ seconds walks up the escalator, she travels $d+\frac{2}{3}d=\frac53d$
$\frac53d$ in $60$ seconds
$=5d$ in $180$ seconds
$=d$ in $36$ seconds
Using algebra and the speed-distance-time relationship
Escalator travels with speed $a$
Aimee walks with speed $b$
When Amy walks up the moving escalator, she has speed $a+b.$
distance = speed $\times$ time, so the length of escalator is given in 3 ways: $$60a=90b=(a+b)t$$
Aim: find $t$ (which might involve also finding $a$ and $b$)
Notice we have $(a+b)\times t$ so could get $t$ by dividing by $(a+b)$
$60a=(a+b)t$ so $180a=3(a+b)t$
$90b=(a+b)t$ so $180b=2(a+b)t$