I have a function $f(x,y,z)$. Can this be rewritten as $f(y,z)$ if $x$ is not present in the function? Let's say that I have $f(x,y,z)$ which is derived from another function g. And this is true for all cases.
For a certain case, $f(x,y,z) = yz$
Is this the same as writing/ defining the function as $f(y,z) = yz$  ?
Or is it imperative that I write it as the $f(x,y,z)=yz$   as its definition?
 A: $
\newcommand{\R}{\mathbb{R}}
\newcommand{\ff}{\widetilde{f}}
\newcommand{\dom}{\operatorname{dom}}
\newcommand{\cod}{\operatorname{cod}}
$
You can't, because it goes back to the definition of the function. The fact that you have $f(x,y,z)$ rather than $f(y,z)$ means $f$ is a function with domain $\R^3$.
(Or, in general, the domain of $f$ is $A \times B \times C$ where $A$ is the set in which $x$ comes from, $B$ is that of $y$, and $C$ is that of $z$, but, for simplicity, let's work in vector spaces over $\R$.)
On the other hand, if you have a function $\ff(y,z)$, this means that the function has domain $\R^2$.
This matters because we say two functions $g,h$ are equal if and only if...

*

*They have the same domain (i.e. $\dom(g) = \dom(h)$)

*They have the same codomain (i.e. $\cod(g) = \cod(h)$}

*For every element in the domain, each function (i.e. $g(x) = h(x)$ for all $x$ in $\dom(g)$ or, same difference, $\dom(h)$)

In this case, $f$ in this discussion has $\dom(f) = \R^3$ and $\ff$ has $\dom(\ff) = \R^2$ -- even if the former does absolutely nothing with the $x$ component of its vector, it's still sending ordered triples or vectors $(x,y,z)$ as a whole somewhere.
So it doesn't make sense, really, to say you can write $f(x,y,z)$ or $f(y,z)$ interchangeably when $f$ does not depend on $x$. One represents a function with domain $\R^3$ and the other one with $\R^2$. They're not really the same thing.
To take it to the extreme, I could define a function $F : X^n \to Y \ne \varnothing$ by
$$F \big( (x_1,x_2,\cdots,x_n \big) = y$$
for some fixed $y \in Y$. Sure, this is a constant map and does not depend on any of the $x_i$. However, to say you could just "ignore" the $x_i$ and write $F(\varnothing) = y$, just begins to look silly. You're still sending stuff somewhere, it just happens to be a place that doesn't depend on the $n$-tuple.

Luckily, this is easily handled, as mentioned in the comments. So you have the function
$$f(x,y,z) = yz$$
and want to define a function purely in terms of $y,z$ for whatever reason. Simply define a new function! That being, you could say something like this:

Define the function $\ff$ by $\ff(y,z) = f(x,y,z)$, i.e. $\ff(y,z) = yz$.

