What exactly is an implicitly defined function?. So I've just started to get into some calculus and I recently came across the topic of implicit differentiation.
I am extremely confused on what implicit functions are and there is very little information on what exactly implicitly  defined functions really are but instead lots of defintions on how to implicitly differentiate.
I became even more confused from this definition  proofwiki provides  which defines an implicit function as follows.
Consider a (real) function of two independent variables $z=f(x,y).$
Let a relation between $x$ and $y$ be expressed in the form $f(x,y)=0$ defined on some interval $I$.
If there exists a function:
$y=g(x)$
defined on $I$ such that:
$∀x∈I:f(x,g(x))=0$
then the relation $f(x,y)=0$ defines $y$ as an implicit function of $x$.
I can't seem to wrap this definition i am familiar with how a function is a special kind of binary relation but when we state that let a relation between $ x, y$ be expressed in the form $f(x,y)=0$ what exactly do we mean by this for example how is $x^2 +y^2=1$ a relation in that form.
a simple example would be greatly appreciated.
I hope this question is not too vague to be answered or too silly to be even considered thanks in advance.
 A: The main idea is that you have an equation that in theory could be solved for $y$ as a function of $x$ but the actual procedure is too hard/unknown.  In other words, the equation gives enough information to uniquely specify a function but doesn't actually give it.
For example, $x^3y^3 + xy = 4$ is an equation cubic in $y$.  With a lot of tedious work in algebra, it is possible to solve it and get $y$ alone.  But if the equation is fifth degree or higher, then depending on the coefficients, there might not be any known way to find the formula directly.
It's a kind of specifying an answer by giving a problem that it solves.  We call it indirect because it isn't "$y = $(blah that includes $x$'s but doesn't include any $y$'s)".
A: Here is what I think the story is.
Step (i) Suppose we have two sets $X,Y\subseteq\mathbb{R}$  and a function $f:X\times Y\to\mathbb{R}$.
Step (ii) This gives us a relation on $X\times Y$: define $x\sim y$ to mean $f(x,y)=0$.
[Note: In fact there's no need to introduce the idea of this relation.]
Step (iii) Suppose that we are in this situation, and that on some interval $I\subseteq X$ we have a function $g:I\to\mathbb{R}$ such that for all $x\in I$ it is true that $f(x,g(x))=0$. Then we say that $f$ defines $g$ implicitly on $I$.
[Note: We sometimes then talk about $g$ being an implicit function - this is loose, informal, talk. More carefully, $g$ is a function, which is implicitly defined on $I$ by $f$.]
A: There are already excellent answers, e.g. RobertTheTutor's that, in summary, say: the equation $f(x,y)=0$ defines a function $y=g(x)$ by solving for $y$ in theory, although in practice it is difficult / impossible / inconvenient to solve.
As per the comment my confusion arises with why can we say that equations which clearly cannot be written explicitly in terms of $$ then be considered equations of $$ in terms of  and also the OP there is very little information on what exactly implicitly defined functions really are but instead lots of definitions on how to implicitly differentiate, here is my interpretation.
The definition of implicit function does not mention the derivative. But it turns out that the most useful way to prove that such implicit function exists, is the implicit function theorem which does use Calculus. There are excellent rigourous proofs in Calculus books but, informally, think that if that function $f$ exists and it is differentiable, then it should be possible to differentiate implicitly the equation $f(x,y)=0$ and solve for $y'$, that necessarily must give $y'=g'(x)$.
So here you are your answer: even if the equation cannot be written  in practice explicitly in terms of $$, it can be considered in theory that it defines a function of $y$ in terms of $x$, if the conditions of the IFT hold, which can be checked by computing the implicit derivative.
EDIT: This question reminds us the very important fact that the IFT is local, i.e. it only asserts the existence of the function implicitly defined in a neighbourhood of some point where you have computed the implicit derivative.
EDIT 2: I think an important source of confussion here is the difference between what absolutely cannot be done and what I cannot do because it is not simple or immediate for me. Imagine I have not yet studied square roots. Then $y^2=x$ cannot be solved in terms of simple (for me) functions. But if a draw this curve near $(x,y)=(1,1)$, it turns out that it "looks like" a function (the vertical line test holds) so I can define ex novo the function $f$ such that its graph matches the curve $y^2=x$. And then, voilà, I am magically able to solve for $y=f(x)$! In other words, $f$ is defined as: given $x$, draw a vertical line that intersects the curve at a single point, which is by definition $y=f(x)$.
