Consider the smooth function $f : \mathbb{R}^2 \to \mathbb{R}$. Applying the chain rule yields
$$\mathrm{d}f = \frac{\partial f}{\partial u} \mathrm{d}u + \frac{\partial f}{\partial v} \mathrm{d}v.$$
This makes perfect sense to me for the independent variables $u$ and $v$. However let $u = x(t)$ and let $v = y(t)$. Then how does it the partial derivative
$$\frac{\partial f}{\partial u}$$
make any sense. Because if $u = x(t)$ is varied, $v = y(t)$ cannot be held constant. Or is $v$ allowed to be any function when taking the partial derivative with respect to $u$, i.e.
$$\left.\frac{\partial f}{\partial u}\right|_{u = x(t)}?$$
When $u=x(t)$ and $v=y(t)$ then you cannot consider $u$ and $v$ independent anymore. In such case, according to the chain rule, the derivative of the composite function $$ g(t)=f(x(t),y(t)) $$ is given by $$ g'(t)=\frac{\partial f}{\partial u}x'(t)+\frac{\partial f}{\partial v}y'(t) $$ where the partial derivatives of $f$ are calculated at the point $(x(t),y(t))$. That is, you compute them normally, assuming that $u$ and $v$ are your independent variables, and then you evaluate at $(x(t),y(t))$.