By differentiating implicitly with respect to $x$ both sides of the implicit equation
$$
\begin{equation*}
2x^{2}+y^{2}=33,\tag{1}
\end{equation*}
$$
since the derivatives of both sides should be equal we get successively:
$$
\begin{eqnarray*}
&&\frac{d}{dx}\left( 2x^{2}+y^{2}\right) =\frac{d}{dx}\left( 33\right) \\
&\Rightarrow &\frac{d}{dx}\left( 2x^{2}+y^{2}\right) =0 \\
&\Leftrightarrow &\frac{d}{dx}\left( 2x^{2}\right) +\frac{d}{dx}\left(
y^{2}\right) =0 \\
&\Leftrightarrow &4x+2y\frac{dy}{dx}=0,\qquad \frac{d}{dx}\left(
y^{2}\right) =2y\frac{dy}{dx}\text{ by the chain rule} \\
&\Leftrightarrow &\frac{dy}{dx}=-\frac{4x}{2y}=-\frac{2x}{y}\tag{2} \\
&\Rightarrow &\left. \frac{dy}{dx}\right\vert _{x=2,y=5}=-\frac{4}{5}.\tag{3}
\end{eqnarray*}
$$
The equation of the tangent line at $(2,5)$ is
$$
\begin{equation*}
y-5=-\frac{4}{5}(x-2),\tag{4}
\end{equation*}
$$
while the equation of the normal line to the curve $2x^{2}+y^{2}=33$ at $(2,5)$ is
$$
\begin{equation*}
y-5=\frac{5}{4}(x-2)\Leftrightarrow y=\frac{5}{4}x+\frac{5}{2},\tag{5}
\end{equation*}
$$
because the slope $m$ of the tangent line and the slope $m^{\prime }$ of the normal line are related by $mm^{\prime }=-1$.
ADDED. In a more general case, when we have a differentiable implicit
function $F(x,y)=0$, let $y=f(x)$ denote the function such that $F(x,f(x))\equiv 0\quad$ ($f(x)$ does not need to be explicitly known). If we differentiate both sides of $F(x,y)=0$ and apply the chain rule, we get the following total derivative with respect to $x$:
$$\frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{
\partial F}{\partial y}\frac{dy}{dx}\equiv 0.\tag{A}$$
Solving $(\mathrm{A})$ for $\frac{dy}{dx}$, gives us the following formula
$$\frac{dy}{dx}=-\frac{\partial F}{\partial x}/\frac{
\partial F}{\partial y}.\tag{B}$$
Best Answer
The normal line is perpendicular to the tangent line. Therefore, its slope is the negative reciprocal of the tangent line.
To find the slope of the tangent line at the point where $x = 1/2$, you evaluate the derivative of the equation $y = 2(x - 1)^3$ when $x = 1/2$.
As you found, the derivative is
$$y' = 6(x - 1)^2$$
Hence, when $x = 1/2$, the tangent line has slope
$$y' = 6\left(\frac{1}{2} - 1\right)^2 = 6\left(-\frac{1}{2}\right)^2 = 6\left(\frac{1}{4}\right) = \frac{3}{2}$$
Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, the normal line has slope $-2/3$.
When $x = 1/2$,
$$y = 2(x - 1)^3 = 2\left(\frac{1}{2} - 1\right)^3 = 2\left(-\frac{1}{2}\right)^3 = 2\left(-\frac{1}{8}\right) = -\frac{1}{4}$$
so the tangent line and normal line both pass through the point $(1/2, -1/4)$.
If you use the point-slope form of the equation of a line
$$y - y_0 = m(x - x_0)$$
you obtain
$$y + \frac{1}{4} = -\frac{2}{3}\left(x - \frac{1}{2}\right)$$
for the equation of the normal line.