You use the normal rules for substitution:
If ##v(z,t)=z/t##, then ##v(z+Δz,t) = (z+Δz)/t##
i.e every time you see a z, you replace it with z+Δz
So you have the LHS limit = 1/t - compare that with the partial derivative.
(Do you know how to take a partial derivative?)
How does this get me to the partial derivative of with respect to z and how would I know that doing all this wouldn't give me a partial with respect to t instead?
The partial derivative of ##v(z,t)## with respect to ##t## would be:
$$\frac{\partial}{\partial t}v(z,t)=\lim_{\Delta t\rightarrow 0}\frac{v(z,t+\Delta t)-v(z,t)}{\Delta t}$$
In general,
If you have a function ##f(\vec{x})## of ##N## variables ##\vec{x}=(x_1,x_2,\cdots ,x_N)##
Then the partial derivative wrt the nth variable, where ##1\leq n \leq N##, is:
$$\frac{\partial}{\partial x_n}f(\vec{x})=\lim_{\Delta x_n\rightarrow 0}\frac{f(\cdots, x_n+\Delta x_n,\cdots)-f(\cdots,x_n,\cdots)}{\Delta x_n}$$
... note: if you don't understand that notation you have a serious problem!
Also, was (z/t) arbitrary? Or would something like (zt) have worked?
Do it and see :)
The definition works for any differentiable function. It would have worked for ##v(z,t)=z## or ##v(z,t)=t## ... whatever.
Do you know how to differentiate functions by rule instead of using the definition?
i.e. can you do: $$\frac{d}{dx}(ax^2+bx+e^{cx})$$