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For a function $f = f(x, y, t)$, and $x = x(t)$, $y = y(x, t)$, according to the chain rule, we have $$ \frac{\mathrm{d}f}{\mathrm{d}t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial t} $$ and for the term $\partial y / \partial t$, it should be expanded as $$ \frac{\partial y}{\partial t} = \frac{\partial y}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial y}{\partial t} \frac{\partial t}{\partial t} $$

Obviously these 2 $\partial y / \partial t$ s have different meanings, while sharing the same symbol. This makes a confusion.

Furthermore, if we have $g(u, v, t) = u^2 + v^3 + t^4$ and $u = u(t, s)$, $v = v(t, s)$, how should I express $\partial g / \partial t = 4t^3$ and do not get confused with $$ \frac{\partial g}{\partial t} = \frac{\partial g}{\partial u} \frac{\partial u}{\partial t} + \cdots$$

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You haven't used the Multivariable Chain Rule correctly. What we should have is

$$\frac{\mathrm{d}f}{\mathrm{d}t} = \frac{\partial f}{\partial y} \frac{\partial y}{\partial x}\frac{\partial x}{\partial t} +\frac{\partial f}{\partial y} \frac{\partial y}{\partial t}+\frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial t} $$
Also, if you get to a point along the chain where you get $\frac{\partial t}{\partial t}$ you have gone too far. Regardless, this fraction clearly gives $1$, so we can ignore it in the multiplication.

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    What I actually want to ask is, in $g(u, v, t)$, the variable $t$ is both explicit and implicit. So when we use $\partial g / \partial t$, we are mentioning the direct result of partial derivative (i.e. $4t^3$ for function $g$), or the result of chain rule? – 2017-02-10