Consider the following paragraph and equations from the NCERT textbook regarding the derivatives of functions in parametric form.
Sometimes the relation between two variables is neither explicit nor implicit, but some link of a third variable with each of the two variables, separately, establishes a relation between the first two variables. In such a situation, we say that the relation between them is expressed via a third variable. The third variable is called the parameter. More precisely, a relation expressed between two variables x and y in the form $x = f (t), y = g (t)$ is said to be parametric form with t as a parameter.
In order to find derivative of function in such form, we have by chain rule.
$$\dfrac{dy}{dt} = \dfrac{dy}{dx}.\dfrac{dx}{dt} $$
$$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \text{(whenever} \dfrac{dx}{dt} \ne 0)$$
I have doubt with the first equation of the chain rule $\dfrac{dy}{dt} = \dfrac{dy}{dx}.\dfrac{dx}{dt} $. As we can see that $y$ is not directly dependent on $x$, but only on $t$, then how can chain rule assumes $\dfrac{dy}{dx}$ in the equation?
Since $x$ and $y$ are both parameterised by $t$, they can't vary independently - they are coupled to each other. Which means that, given some value of $x$, there are only very specific values that $y$ can take.
In fact, as long as we are careful about things, we can say that since $x$ is a function of $t$, you can invert that so that $t$ is a function of $x$ - i.e. $x = f(t) \implies t = f^{-1}(x) = t_x(x)$. Then that means that $y$ can also be written as a function of $x$, ie. $y = g(t_x(x))$, and under a few simple assumptions you can say that differentiablity is preserved when you do so.