In the formula description there is one extra partial derivative compared to the example solution. What's the difference here? What's the physical implication of the last partial derivative in the description when we're already taking the derivative of each variable? Isn't it unnecessary?
What is confusing here is that the same function name $f$ is used to denote two different functions (this is a commonly used convention which does not cause any confusion, when it is understood what is behind).
Let us take a function $f(x,y,z,t)$, which is a function of four variables $x,y,z,t$. Let is now assume that $x(t), y(t), z(t)$ are also functions - functions of time. We can then define function $$F(t) = f(x(t),y(t),z(t),t),$$ which is a function of time only. Evaluating the derivative of this function using the chain rule of differentiation we obtain the first equation in the question.
The partial derivative with respect to $t$ to me only occured because there was an explicit time dependence in the function $f$, think of it like this, $x$, $y$, $z$ are all functions of $t$ so $f$ as a whole is also a function of $t$, think of it by an example, $f$ gives the particle number density in a flowing fluid, imagine a small blob of this fluid, as time passes, this blob will move, it moves from point $a$ to $b$ in time $t$, now imagine that this blob goes from $a$ to $b$ (position vector changes by $dr$) instantaneously, the change in density will be $\frac{\partial f}{\partial r}dr$, now time starts running and the density at $b$ changes by $\frac{\partial f}{\partial t}dt$ (it is a continuous process really I just broke it down into Subprocesses for understanding purposes) total change will be $\frac{\partial f}{\partial r}dr+\frac{\partial f}{\partial t}dt$, now divide by $dt$ on both sides (as the quantities $dt$ and $dr$ are very small this is the derivative of $f$ wrt $t$) you will get the desired results, if $f$ had no explicit time dependence the second increment would. Not have happened
