$\mid {9y-6} \mid =3$
squaring both side
$81y^{2}-108y+36=9$
dividing both side by $9$
$9y^{2}-12y+4=1$
again dividing by $9$
$y^{2}-\dfrac{4}{3}y+\dfrac{4}{9}=\dfrac{1}{9}$
$\Rightarrow y^{2}-\dfrac{4}{3}y=\dfrac{1}{9}-\dfrac{4}{9}=-\dfrac{1}{3}.$
Hence answer is option $C$