The interval, in which function $y=x^3+6x^2+6$ is increasing is: 

Given: $y=x^3+6x^2+6$
$\Rightarrow f^{\mathrm{\prime }}\left(x\right)=\frac{dy}{dx}=3x^2+12x=3x(x+4)$
Taking $f^{\mathrm{\prime }}\left(x\right)=0,$ we get, $x=0\mathrm{\&}x=-4$
Taking $f^{\mathrm{\prime }}\left(x\right)>0$ $\text{ for }x<-4\text{ and }x>0$
$\text{ And, }{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)<0\text{ for }-4<x<0\text{. }$
Hence, $f\left(x\right)$ increases on the interval $(-\mathrm{\infty },-4)\cup (0,\mathrm{\infty })$