(ii) $y=18 x-107$

(i) Method 1
$$
y^3=4-\frac{x y^2}{2}
$$
Differentiate with respect to $x$ :
$$
\begin{aligned}
3 y^2 \frac{\mathrm{d} y}{\mathrm{~d} x} &=-\frac{y^2}{2}-x y \frac{\mathrm{d} y}{\mathrm{~d} x} \
\left(3 y^2+x y\right) \frac{\mathrm{d} y}{\mathrm{~d} x} &=-\frac{y^2}{2} \
\frac{\mathrm{d} y}{\mathrm{~d} x} &=-\frac{y^2}{6 y^2+2 x y}=-\frac{y}{6 y+2 x} \text { (shown) }
\end{aligned}
$$

(ii)

At $y=1, x=6$ and $\frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{1}{18}$
Gradient of normal $=18$
Hence equation of normal
$y-1=18(x-6)$
$y=18 x-107$