It turns out there are two separate issues to consider.
In functional notation, derivatives are things that are applied to functions, not variables. The derivative of a univariate function (i.e. a function with one argument) is always the derivative of the value of the function with respect to the argument of the function.
i.e. if $f$ is the function defined by $f(x) = x^2$, then $f'$ is the function defined by $f'(z) = 2z$.
In the equations above, $x$ and $z$ are dummy variables; they have no meaning on their own, and only purpose in existence is to let us write down an equation for the value of $f$ at a point.
In dependent variable notation, the variables you use all have some intrinsic meaning. (e.g. you might use $t$ to refer to "time"). You can't differentiate variables, but you can take their differentials. The differential of $x$ is $dx$. The differential of $x^2$ is $d(x^2) = 2x~dx$.
Sometimes, two differentials can be proportional. For example, if $x$ and $t$ are dependent one one either via the equation $x = t^2 + 1$, then this equation also holds when we compute the differential on both sides: $dx = 2t~dt$.
In Leibniz notation, when we have such a proportion, we use $dx/dt$ to express the ratio. So if $dx = 2t~dt$, then we say $dx/dt = 2t$. And $dt/dx = 1/(2t)$.
If the relationship between $x$ and $t$ is $x = f(t)$, then fortunately we have $dx = f'(t) dt$, and so in Leibniz notation, $dx/dt = f'(t)$.
If we have two equations, such as
$$ \frac{8.5}{10-x} = \frac{1.5}{y} $$
and
$$ x = 2.2t $$
then we can get two equations between the differentials. Let me first simplify the first equation to
$$ \frac{10-x}{8.5} = \frac{y}{1.5} $$
Now, when we take the differential, we get two equations
$$ dx = 2.2~dt $$
$$ -\frac{1}{8.5} dx = \frac{1}{1.5} dy $$
and if we wanted, we can solve the first for $dx$ and plug it into the second:
$$ -\frac{2.2}{8.5} dt = \frac{1}{1.5} dy $$
We can't always write differentials as proportions. e.g. if $A = xy$, then $dA = x dy + y dx$. If $x$ and $y$ aren't functionally related to each other, then $dA/dx$ and $dA/dy$ simply don't make sense.
Best Answer
The point is we want to say a curve is smooth, for example $C^1$ (continuously differentiable), if it has a smooth parametrization. And without that condition on non-vanishing derivatives that would admit "smooth" curves that we really don't want to call "smooth".
Example: Consider the curve in the plane defined by $y=|x|$. That has a corner at the origin - our definition of "$C^1$ curve" had better not include this curve, right?
But define $f,g:\Bbb R\to\Bbb R$ by $g(t)=t^2$ and $$f(t)=\begin{cases} t^2,&(t\ge0), \\-t^2,&(t<0).\end{cases}$$Then $f$ and $g$ are both $C^1$, and $x=f(t)$, $y=g(t)$ is a parametrization of the curve $y=|x|$. Without the condition on non-vanishing derivatives this would make $y=|x|$ a $C^1$ curve. (With the condition on non-vanishing derivatives everything's fine - this parametization is disallowed because $f'(0)=g'(0)=0$.)