How to take parametric equations (x, y) to create a derivative formula? I always thought that if I take the derivative of the y and x equation and divide y' by x', then that would be the derivative in formula form.
Is this correct?
 A: Yes, in many cases you may compute $$y'(t) = \frac{dy}{dt}\;\;\text{ and}\;\;x'(t) = \frac{dx}{dt} \quad \text{when}\;\;x'(t)\neq 0; \text{ and not both} \; x' = y' = 0$$
Then $$ \dfrac{dy}{dx} = \dfrac{y'(t)}{x'(t)}= \dfrac{dy/dt}{dx/dt}$$
But note that you will need to do more work if $\dfrac{dy}{dx}$ at some given $t$(s) evaluates in the form $\dfrac 00$, i.e., when both $x'(t) = y'(t) = 0$, for given $t$.
A: Your intuition is essentially correct, although certain parametrizations "come to a stop," so $\frac{dy}{dx}$ is of the form $\frac{0}{0}$.
Consider, for example, the parametrized curve
$$
\begin{cases}
x(t) = (t - 1)^3 \\
y(t) = (t - 1)^6.
\end{cases}
$$
It's clear that $y = x^2$, and
$$
\begin{cases}
x'(t) = 3(t - 1)^2 \\
y'(t) = 6(t - 1)^5,
\end{cases}
$$
so
$$
\frac{~\frac{dy}{dt}~}{\frac{dx}{dt}} = \frac{6(t - 1)^5}{3(t - 1)^2} = 2(t - 1)^3 \quad \text{only for } t \ne 1.
$$
But if we imagine $t$ measuring time, then $(x(t), y(t))$ is the position of a point at time $t$.  The point traces out the curve $y = x^2$, but in such a way that it slows to a complete stop at $t = 1$ (notice that $(x'(1), y'(1)) = (0, 0)$), only to resume moving afterwards.
What direction is the particle moving at $t = 1$?  (It's a trick question:  the particle's not moving.)  But we want to say that it's moving horizontally, or equivalently, in a direction of slope $\frac{dy}{dx} = 0$.
In a sense, that's correct, since we can the limit.
$$
\frac{dy}{dx}\Bigg|_{(x, y) = (1, 1)} = \lim_{t \to 1} \frac{~\frac{dy}{dt}~}{\frac{dx}{dt}} = 0.
$$
A: From the chain rule, we have
$$\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt} \implies \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$$
Hence, what you have is correct from the chain rule, provided $dy/dt$ and $dx/dt$ both exist. Note that we also need $dx/dt$ to be non-zero.
