The values of x for which $x^3-6x^2-36x+7$ increases are

$y=x^3-6x^2-36x+7$

$\Rightarrow \frac{dy}{dx}=3x^2-12x-36>0$

$3x-12x-36>0\Rightarrow 3(x^2-4x-12)>0$

$\Rightarrow (x+2)(x-6)>0\Rightarrow x<-2\left(or\right)6<x$

The range of values of x is 

$(-\infty ,-2)\cup (6,\infty )$