Can any curve in $\Bbb R^2$ (which doesn't intersect itself) be a graph of a certain equation?
In other words, if given an arbitrary curve in $\Bbb R^2$ (which doesn't intersect itself), is there a equation $f(x,y)=0$ that takes the given curve as its graph?
Edit: I was a bit confused about the definition of 'function'. What I meant was 'any equation $f(x,y)$' , such as $x^2+y^2=1$; and should not be broken down into several intervals. Sorry for the mistake.
Edit: I was asking about a continuous(?) function, as GEdgar said in the comments. I couldn't think of the right word...

I think what you're asking is this:
Given a set $S \subset \mathbb{R}^2$ that represents a curve, is it always possible to find a function $f:\mathbb{R}^2 \to \mathbb{R}$ such that $S = \{(x,y) \in \mathbb{R}^2 : f(x,y)=0\}$.
The problem is that we don't have a clear definition of "curve", so we don't know much about the properties of the set $S$.
However, with some suitable restrictions on $S$, we can define a function $f$ by $$ f(x,y) = \text{(minimum) distance from } (x,y) \text{ to } S $$ This will often give you the familiar equations of various curves (like lines and circles, for example).
On the other hand, no matter what set $S$ you're given, you can define a function $f$ as suggested in Rahul's comment $$ f(x,y) = \cases{0 \quad\text{ if } (x,y) \in S \\ 1 \quad\text{ if } (x,y) \notin S } $$
Then it's certainly true that $S = \{(x,y) \in \mathbb{R}^2 : f(x,y)=0\}$, as you wanted, but the argument is somewhat circular, and we still don't know what "curve" means.