Question: I asked 3 professors at my university and none gave me a clear cut answer, but instead merely told me qualities of this notation.
Here is what I understand so far from what they told me:   

1)Treat the top variable as as variable when finding the derivative  

2)Treat the bottom variable as a constant when finding derivative  

3)It it said "Find x with respect to y", but what exactly does that mean? What does it mean for something to be in respect to something else?

It seems like $\frac{dx}{dy}$ notation changes according to values in the problem. For instance, If $y = x^3 + 2x$ and $\frac{dx}{dt} = 5$, find $\frac{dy}{dt}$ when $x=2$. Why do the values of $\frac{dx}{dy}$ change in this problem and how do I solve this?

Answer: 
$\frac{d}{dx}$, $\frac{d}{dt}$, $\frac{d}{dy}$, etc. are all linear differential operators. They specifically mean "differentiate the thing that comes after with respect to whatever the variable on the bottom is." The notation is sometimes problematic, because it is easy to abuse.

When we write something like $\frac{dx}{dt}$, what we're really saying is "differentiate $x$ with respect to $t$." This statement is, of course, meaningless unless $x$ has some dependence on $t$.

For instance, let's say that $x = at^2+bt+c$. Then, $\frac{dx}{dt}$ is basically saying "differentiate $at^2+bt+c$ with respect to $t$. But since that polynomial is equal to $x$, let's just write it as $\frac{dx}{dt}$."

When we apply $\frac{d}{dt}$ to $x$, we're not multiplying by some quantity $\frac{d}{dt}$. In fact, $\frac{d}{dt}$ is not a quantity at all, but rather an operation. The way we write it is tricky, because it can be confusing.

For more details on why we write it in this way, see the excellent answer here: <https://math.stackexchange.com/a/21209/31475>
$\frac{d}{dx}y(x)$ denotes how $y$ varies according to $x$, and viceversa. It is only a metter of notation, even though there's a lot more to know.

Moreover, in your last example:
$$
\frac{dx}{dt}=5\Rightarrow x(t)=5t+x(0).
$$
So:
$$
\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}=(3\cdot 2^2+2)\cdot 5.
$$