Under what condition can I confidently do $(dx/dt)/(dy/dt)= \frac{dx}{dy}$ This question has been bothering me for a while now
Can someone provide a necessary condition for relation
$$\frac{\frac{dx}{dt}}{ \frac{dy}{dt}} = \frac{dx}{dy}$$
to hold true where $x$, $y$ are functions of $t$
 A: It's certainly necessary
for $x$ and $y$ to be differentiable functions of $t$ and $dy/dt \ne 0$, otherwise your relation 
doesn't make sense.  Using the Implicit Function Theorem, it's sufficient 
for $x$ and $y$ to be continuously differentiable and $dy/dt \ne 0$.  Presumably the continuity of the derivatives is not necessary, but I don't know if there's a simple necessary and sufficient condition.
EDIT: Basically the problem here is to have $x = X(t)$ and $y = Y(t)$ define $x$ as a function of $y$ in a neighbourhood, say, of $y = Y(t_0)$.  Once you have that, 
$\dfrac{dx}{dt} = \dfrac{dx}{dy} \dfrac{dy}{dt}$ is the chain rule.  To see that it's not sufficient for $X$ and $Y$ to be differentiable functions of $t$ with $Y'(t_0) \ne 0$, consider e.g.
$$  \eqalign{Y(t) &= t + t^2 \sin(1/t) \ \text{for}\ t \ne 0\cr
Y(0) &= 0\cr
X(t) &= t\cr}$$
where $Y$ is not a one-to-one function in any neighbourhood of $t_0$.
 On the other hand, it's not necessary for $Y'(t) \ne 0$ in a neighbourhood of $t_0$: it's possible to modify this example to have $Y'(t_n) = 0$ on a sequence of points $t_n$ approaching $0$, but $Y(t)$ one-to-one near $0$.
