Let f be a differentiable function in x, y, and z (which is stronger than just saying that the partial derivatives exist). Further, suppose x, y, and z are differentiable functions of some parameter t. Then f(t)= f(x(t), y(t), z(t)) describes f values along a specific smooth path in R3. If you like, you can think of it as the trajectory of an object in space.
By the chain rule,
[tex]\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{df}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}+ \frac{\partial f}{\partial z}\frac{dz}{dt}[/tex]
Since that is an ordinary derivative of a function of a single variable, we can define its "differential":
[tex]df= \frac{df}{dt} dt= \left(\frac{\partial f}{\partial x}\frac{df}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}+ \frac{\partial f}{\partial z}\frac{dz}{dt}\right)dt[/tex]
Or course, that is the same as
[tex]df= \frac{df}{dt} dt= \frac{\partial f}{\partial x}\frac{df}{dt}dt+ \frac{\partial f}{\partial y}\frac{dy}{dt}dt+ \frac{\partial f}{\partial z}\frac{dz}{dt}dt[/tex]
[tex]df= \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+ \frac{\partial f}{\partial z}dz[/tex]
where all mention of "t" has disappeared.