Given curve is $y^{3}-x y-8=0$
On differentiating w.r.t. $x$, we get
$ 3 y^{2} \frac{d y}{d x}-x \frac{d y}{d x}-y-0=0 $
$ \Rightarrow \frac{d y}{d x}\left(3 y^{2}-x\right)=y$
$\frac{d y}{d x}=\frac{y}{\left(3 y^{2}-x\right)} $
$ \Rightarrow \left(\frac{d y}{d x}\right)_{(0,2)}=\frac{2}{3(2)^{2}-0}=\frac{1}{6} $
$\therefore $ The slope of the normal $=-\frac{1}{d y / d x}=-6$