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
$$\frac{dy}{dx}=-\frac{\partial F}{\partial x}/\frac{
\partial F}{\partial y}.\tag{B}$$
$x$ and $y$ are functions of $t$, so it would be more appropriate to write $z$ as
$$z=\sqrt{[x(t)]^2+[y(t)]^2}$$
Using the chain rule for $t$,
$$\frac{\mathrm{d}z}{\mathrm{d}t}=\frac{\partial z}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial z}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}$$
$\frac{\mathrm{d}x}{\mathrm{d}t}$ and $\frac{\mathrm{d}y}{\mathrm{d}t}$ are already known, so all you have to do is find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$.
I understand that we need the derivative of the distance function, but what I don't grasp is: the derivative with respect to what?
In the end, you're taking the derivative with respect to time, $t$. This becomes clearer if you simply write
$$z=f(x,y)$$
instead of
$$z^2=g(x,y)$$
as you have done.
The derivative of $z^2$ is exactly $2z$ (by power rule), so why are we multiplying by $\frac{dz}{dt}$ too?
I would avoid doing it like you did, to avoid implicit differentiation.
Best Answer
The square distance $d$ helicopter - soldier is: $$d^2=(x^2+7-7)^2+(x-3)^2$$ The minimum of this distance is given when the derivative of $d^2$ is zero. So: $$(d^2)'=4x^3+2x-6$$ which is zero for $x=1$ Putting $x=1$ in the equation of the helicopter trajectory, you have: $$P_m=(1,8)$$ which is the helicopter coordinates in wich the distance is minimum.