How do you formally prove that a function in several variable is really a function Let say for example that we define $f:\mathbb{R}^{3}\longrightarrow \mathbb{R}^{3}$ such that $f(x,y,z)=(y^{2},xz,xy^{2})$. My informal argument would be just that there is only one object  that can be defined having three real numbers, but I'm just saying this intuitively. How do you prove it formally?
 A: Let $P=(x_1,y_1,z_1)=(x_2,y_2,z_2)=Q$ then $$x_1=x_2,~~y_1=y_2,~~z_1=z_2$$ then $$y_1^2=y_2^2,~~x_1z_1=x_2z_2,~~x_1y_1^2=x_2y_2^2$$ So $f(P)=f(Q)$. This is very elementary way.
A: From the comments, it sounds like you are interested in defining $\mathbb{R}^d$. Afterall, Dedekind cuts are used in constructing the real numbers. If you've defined $\mathbb{R}$ via your favorite axioms (say through Dedekind cuts), then $\mathbb{R}^d$ is defined as all ordered $d$-tuples whose elements take values in $\mathbb{R}$. In other words, say $d=3$ then $(3,1,0.1)=(3,1,0.1)$ but $(3,1,0.1)\neq (0.1,3,1)$. 
By definition, $f:\mathbb{R}^d\rightarrow\mathbb{R}^d$ maps every point in $\mathbb{R}^d$ to some other point in $\mathbb{R}^d$. In general, the definition of a function is something that maps one set to another set with each input having one output. So in your example, $f(x,y,z)=(y^2,xz,xy^2)$, is perfectly well defined, as each component, say $y^2$, is well defined on $\mathbb{R}$. Your function in each component has some rules in particular, multiplication being one of them, and that's well defined in the field of real numbers from whichever construction you used. 
A: Often, it's useful to analyze a given function as a composition of "smaller" elementary functions. Here, we have the three coordinate functions $f_j : \mathbb{R}^3 \rightarrow \mathbb{R}$ ($j=1,2,3$) defined by $f_1(x,y,z)=y^2$, $f_2(x,y,z)=xz$ and $f_3(x,y,z)=xy^2$. You could even break these down further. For instance, $f_3$ is the product of the projection $(x,y,z) \mapsto x$ and the function $(x,y,z)\mapsto y^2$, and the product of two real-valued functions with common domain is also a real-valued function.
Finally, given three real-valued functions $f_1,f_2,f_3$ with common domain $\mathbb{R}^3$, we can form the function $f(x,y,z)=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$ with domain $\mathbb{R}^3$ and codomain $\mathbb{R}^3$.
This example kind of clouds the set theory; here's another example. Suppose $f : A \rightarrow B$ and $g : A \rightarrow C$ are both functions. Note that they share the domain $A$. Then we can form a function $h : A \rightarrow B \times C$ where $B \times C$ is the Cartesian product of the sets $B$ and $C$. The function $h$ is defined in terms of $f$ and $g$ via the formula $h(a)=(f(a),g(a))$ for $a \in A$.
I don't believe this construction has a name. In fact, it can be written as a composition of the diagonal map $a \mapsto (a,a)$ going from $A \rightarrow A \times A$ along with the Cartesian product of the functions $f$ and $g$. Note that here I'm talking about the Cartesian product of two functions.
Finally, to glue all of this thinking together, you need to prove that the composition of two functions is actually a function. It sounds silly, but this type of thinking is useful in more advanced classes: the composition of continuous functions is continuous, the composition of differentiable functions is differentiable, the composition of homomorphisms is a homomorphism, etc.
