What is the relation of gradient and total differential.
If $f(t,x,y,z)$ be a differentiable function such that $x,y,z$ are function of $t$
then total differential $$df=(f_t+f_xx'+f_yy'+f_zz')dt=\color{red}{(f_t+\nabla f.\frac{dr}{dt})dt}$$ where $'$ means $\frac{d}{dt}$.
Is the red part correct? Is it sufficient to show?
You should think about the total differential as a linearisation of sorts. It is precisely the linear map Df from $\mathbb{R}^n \to \mathbb{R}^k$ that approximates the function $f: \mathbb{R}^n \to \mathbb{R}^k$ locally the best (one can prove this more rigorously but I would suggest to google for "Lecture notes on real analysis" in case you want to know the details)
Now, the gradient is a special case of the total differential. In case your codomain is $\mathbb{R}$ you get that the transformation matrix of the total differential – called the Jacobi matrix – is precisely the gradient.
As you stated the primes just mean $\frac{d}{dt}$ so $x'$ corresponds to $\frac{dx}{dt}$ and if you multiply out the inner product you will see that your statement is correct.