The interval in which $y=\frac{1}{4x^3-9x^2+6x}$ is increasing 

$y^{\text{'}}\left(x\right)=-\frac{12x^2-18x+6}{{\left(4x^3-9x^2+6x\right)}^2}$

$=-\frac{6(2x-1)(x-1)}{{\left(4x^3-9x^2+6x\right)}^2}$

$y^{\text{'}}\left(x\right)>0\iff (2x-1)(x-1)<0$

$\iff \text{ (i) }2x-1>0, x-1<0$

$\text{ (ii) }2x-1<0, x-1>0$

$\text{ (i) }\Rightarrow x\in (1/2,1)\text{ case (ii }$ is not possible. Hence y increases on (1/2, 1)