Implicit differentiation problem; find normal line to the given graph PROBLEM:

Heat flows normal to isotherms, curves along which the temperature is
  constant. Find the line along which heat flows through the point
  $(2,5)$ when the isotherm is along the graph of $2x^2+y^2=33$.

QUESTION:
Here it gives me the equation of the graph (original function). I know that by finding the derivative of the expression it gives I can substitute $x$ ($x=2$) in and solve for the slope. Then I can use $y = mx + b$ to get the equation of the line as it asks for.
However, it doesn't show the function in $f(x)$ form. Therefore, I'm told that I need to use Implicit Differentiation before I do anything. This is where I'm completely lost. I have no idea what Implicit Differentiation means and how to get this expression in $f(x)$ form so I can find the derivative. Can someone please guide me. I would appreciate it if you could comment on your steps so I can have a more intuitive understanding.  
 A: 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
$$\frac{dy}{dx}=-\frac{\partial F}{\partial x}/\frac{
\partial F}{\partial y}.\tag{B}$$
A: To differentiate implicitly is easy, you'll see.
Take the equation and differentiate like you usually do, with respect to $x$:
$$2x^2 + y^2 = 33$$
Differentiate $2x^2$ to get $4x$.
Differentiate $y^2$ to get $2y\dfrac{dy}{dx}$. Notice that since there is no $x$ here, but a y, which is dependent on $x$, you add the $\dfrac{dy}{dx}$.
Differentiate $33$ to give zero.
And the you get the derivative that Ron Gordon obtained, that is:
$$4 x+ 2 y \dfrac{dy}{dx} = 0$$
I hope this helps you understanding implicit differentiation!
Additional:
Say you had $y$ instead of $y^2$. You differentiate $y$ to get $$1\dfrac{dy}{dx} = \dfrac{dy}{dx}$$
Similarly: $4y^3$ becomes $$12y^2\dfrac{dy}{dx}$$
A: You want the normal to the isotherm, which is normal to a tangent.  You find the tangent by differentiating:
$$4 x+ 2 y y' = 0 = y' = -2\frac{x}{y}$$
At the point $(2,5)$, that tangent slope is $y' = -4/5$.  This means that the slope of the normal to the isotherm at this point is $5/4$.  (The negative reciprocal)  Now you have a slope and a point - can you find the equation of the line?
