Why it is not a function This question is very basic one but at this moment I could not answer it.
We know $y=\pm\sqrt{x}$ is not a function because one value of $x$ corresponds to more than one value of $y$. Again $y=x^2$ is a function. But we can get the first 
equation from the second one, so the first one should be a function, making contradiction.
Could anyone please tell me what I am missing.
Thanks in advance.
 A: A function is simply a box into which you plug a number, and the box spits out one number.
$$f: x \mapsto x^2$$
is a function. If I feed it the number $4$, it will return $16$ and only $16$. The thing with this function is that it is not injective, i.e., if you feed it $-4$, it will also return $16$. But that does not mean it is not a function: it is still returnin one and only one number for any number I feed into it.
The fact that it is not injective, however, means that it is not invertible.
On the other hand, $$f:x\mapsto \pm \sqrt{x}$$
is not a function because, if I feed it $16$, it will return $\pm 4$, so it will return $+4$ and $-4$. 
$$f_+: x\mapsto \sqrt{x}$$
is a function, and so is
$$f_{-}:x\mapsto-\sqrt{x},$$
nut the whole function does satisfy the conditions to be a function.
A: To expand on @5xum answer a bit and also to make one note:
$ y = x^2 $ is not a function either, it's an equation.  
$x^2$ is also not a function, it's an "expression".  A function takes some input and returns a deterministic output ( eg. unique output for a given input ).  One example is the function that takes $x$ and returns $x^2$, the square of the value that was put in.  
Do not be discouraged by your confusion about functions.  In several years of teaching calculus to new undergraduates, by far the part that confused them the most was understanding what a function was.  In many cases they had seen calculus before and retained very little of it other than some symbolic manipulations, but had no understanding of what those manipulations were doing because they did not understand the function concept.  I honestly believe that really understanding what a function is is the main difference between people who "get" mathematics and those who swear they will never get it.   So keep at it and good luck.
Oh, one more thing.  You will definitely see people, including mathematicians call $ y = x^2 $ a function, but this is an "abuse of language", and if you press them on it, I think they will agree.  So, if your teacher said " $ y = x^2 $ is a function but $ y = \pm \sqrt{x}$ is not " that is neither surprising nor correct.  
A: $y=x^2$ is an equation for (I assume) real variables $x,y$. You can say $y$ is a function of $x$. This is because for every $x\in\mathbb{R}$, we have exactly one $y$ which satisfies the equation.
The equation is equivalent to $x=\pm\sqrt y$, in the sense that $\forall x\in\mathbb{R}, y\in\mathbb{R}_{\ge 0},\quad y=x^2\quad\text{if and only if}\quad x=\pm\sqrt y$.
So given the equation $x=\pm\sqrt y$, you can still say that $y$ is a function of $x$. However, $x$ is not a function of $y$, because there is an $y=4\in\mathbb{R}$ such that we have two possible solutions for $x$, $\{2,-2\}$, which satisfy the equation. To be precise, in the beginning, you need to determine what values of $y$ your function should be defined on, in this case, it must be $\mathbb{R}_{\ge 0}$ (non-negative real numbers), but still, there is the $\pm$, which makes it not possible.
So if you do not want to be confused, do not say "<equation> is a function" nor "$y$ is a function of $x$" (where $y$ and $x$ are variables), but explicitly construct the function by a mapping like $x\mapsto x^2$.
