$$\frac{d z}{ dt}=\frac{\partial z}{\partial x_1}\frac{\partial x_1}{\partial t}+\frac{\partial z}{\partial x_2}\frac{\partial x_2}{\partial t}+\frac{\partial z}{\partial x_3}\frac{\partial x_3}{\partial t}+\frac{\partial z}{\partial x_4}\frac{\partial x_4}{\partial t}$$
But if we set $t=1$, then $f$ looks like $f(1,1,-1,1)$, and so there is no $"t"$, hence the partials $\dfrac{\partial z}{\partial x_1}\dfrac{\partial x_1}{\partial t}=\dfrac{\partial z}{\partial x_1} (0)=0$, and the entire thing would just be $0$, since there is no value of $t$ anywhere?
Is this correct? Probably doesn't seem like it?
Because you are replacing $t=1$ before differentiating. The chain rule gives \begin{align} & \frac{\partial z}{\partial x_1}\frac{dx_1}{dt} +\frac{\partial z}{\partial x_2}\frac{dx_2}{dt}+\frac{\partial z}{\partial x_3}\frac{dx_3}{dt}+\frac{\partial z}{\partial x_4}\frac{dx_4}{dt} = \\ = & \frac{\partial z}{\partial x_1}+2\frac{\partial z}{\partial x_2}t-2\frac{\partial z}{\partial x_3}t +3\frac{\partial z}{\partial x_4}t^2. \end{align} Now setting $t=1$ you get $$\frac{\partial z}{\partial x_1}+2\frac{\partial z}{\partial x_2}-2\frac{\partial z}{\partial x_3} +3\frac{\partial z}{\partial x_4} $$ evaluated at $(1,1,-1,1)$.
Replacing $t=1$ means that you are evaluating $f$ at a point corresponding to a specific value of $t$, getting the point $(1,1,-1,1,f(1,1,-1,1)) \in \mathbb{R}^5$. It does not make any sense to differentiate now. To see why, think about the geometric meaning of derivative: what is the slope of a point$\dots$ ? In general you first differentiate getting a new function, then evaluate it at the given point.