Why is $\frac{\partial x}{\partial y} = \frac{dx/dt}{dy/dt}$ if $x$ and $y$ are unknown? Given $\frac{dx}{dt} = f(x,y,t)$ and $\frac{dy}{dt} = g(x,y,t)$. Suppose the explicit form of $x$ and $y$ are unsolvable, then the phase plane is usually considered by looking at $\frac{dx}{dy}$. This represents how much $x$ varies with respect to $y$, which is also $\frac{\partial x}{\partial y}$. Thus, $$\frac{\partial x}{\partial y} = \frac{dx/dt}{dy/dt}.$$
This means that without knowing $x$ and $y$, I can find the derivative of $x$ with respect to $y$ and vice versa. Is there any problem with this?
In a hand-wavy manner, I am wondering if the last equality in the following makes sense
$$\frac{\partial x}{\partial y} = \frac{\partial x}{\partial t}
\frac{\partial t}{\partial y} = \frac{\partial x/\partial t}{\partial y/\partial t}
= \frac{dx}{dy}.$$
As Lee Mosher points out, the case of $dy/dt=0$ (or even $dy/dt \equiv 0$) needs to be considered separately. For this question, I am most interested in the case when $dy/dt >0$.
 A: The expression
$$
\frac{dx}{dy}=\frac{dx/dt}{dy/dt}
$$
is shorthand for one of two more complicated expressions:
$$
X^\prime(z)=\frac{x^\prime(T(z))}{y^\prime(T(z))}
$$
or
$$
X^\prime(y(t))=\frac{x^\prime(t)}{y^\prime(t)}\ ,
$$
where $\ T\ $ is an inverse of $\ y\ $ over some domain, and $\ X(z)=x(T(z))\ $.  These identities are derived by differentiating the identities
\begin{align}
y(T(z))&=z\ \text{ and}\\
X(z)&=x(T(z))\ ,
\end{align}
with respect to $\ z\ $.  All the functions here are functions of a single variable. While it's strictly speaking not an error to denote the derivative of a function $\ \phi(\theta)\ $ of a single variable $\ \theta\ $ by the expression $\ \frac{\partial\phi}{\partial{\theta}}\ $, I can't see how that would ever serve any useful purpose, and it's apt to be confusing, so I'd strongly suggest not doing so.
The components $\ (x, y)\ $ of any solution to your differential equations are functions of a single variable $\ t\ $.  If you define $\ X\ $ and $\ T\ $ as above you will have
\begin{align}
X^\prime(z)&= \frac{x^\prime(T(z))}{y^\prime(T(z))}\\
&=\frac{f\left(X(z),z, T(z)\right)}{g\left(X(z),z, T(z)\right)}\ .
\end{align}
When either $\ f\ $ or $\ g\ $ depend explicitly on their third argument $\ t\ $, this isn't much help, because you can't normally find out what $\ T\ $ is until you've already obtained the function $\ y\ $. However, if neither $\ f\ $ nor $\ g \ $ depends expicitly on its third argument, then the equation becomes
$$
X'(z)= \frac{f\left(X(z),z\right)}{g\left(X(z),z\right)}\ ,\\
$$
which you can write, using the customary shorthand as
$$
\frac{dx}{dy}=\frac{f(x,y)}{g(x,y)}\ .
$$
Now it may well be possible to integrate this differential equation to find an  expression for $\ x\ $ in terms of $\ y\ $, even when it has not been possible to find an expression for $\ y\ $ in terms of $\ t\ $.
