The slope of normal to the curve y=3x^2-6x at x= 0 is :

$y=3x^2-6x$

differentiating wrt x

$\frac{dy}{dx}=6x-6$ (Slope of tangent)

$\frac{-dx}{dy}=\frac{1}{6-6x}$ (Slope of Normal)

at $x=6$ → $\frac{-dx}{dy}]_{x=0}=\frac{1}{6}$