Why does the square root of an inverse function turn negative? 
For example, 
$$f(x)=x^2$$
$$y=x^2$$
$$-\sqrt{x} = f^{-1}$$
Why does $\sqrt{x}$ become negative?

Edit: Sorry for all the confusion, I will state the problem on my textbook and the solution.
"Find F^{-1} and verify that (f · f^{-1})(x) = (f^{-1} · f)(x) = x"
$$f(x) = x^{2}, x≤0$$ 
Solution: 
$$ x=-\sqrt{y}$$ 
Interchange x and y.
$$ y=-\sqrt{x}$$ 
$$f^{-1} (x) = - \sqrt{x}$$ 
Verify.
For x≥0 
$$(f^{-1} · f)(x) = f(-\sqrt{x}) = (\sqrt{x})^{2} = x$$ 

And all I want to know is why $$y=x^{2}$$ become into $$f^{-1}(x)=-\sqrt.{x}$$
 A: There seems to be a lot of confusion here. Let's remember this very important definition: a function maps any one input to only one output. This means that you can never have $f(x) = 3$ and $f(x) = 5$, because that would mean that $3=5$, which is clearly wrong. As a rule, we have that if $f(x) = a$ and $f(x) = b$ then $a = b$.
So let's take $f(x) = x^2$. You know what that looks like --- it's a parabola. For example, we have that $f(3) = 9$ and $f(-3) = 9$. This means that $f(x)$ is not invertible. Why? If you tried to invert this function, then $f^{-1}(9) = 3$ and $f^{-1}(9) = -3$. But $3 \neq -3$. So $f^{-1}$ violates the rule we laid out above. 
A: The function $f(x)=x^2$ doesn't actually have an inverse function.
Let's see why:
If we follow the standard procedure for calculating the inverse function, we get that the inverse function should be $f^{-1}=\pm\sqrt{x}$. 
Now, we have a problem. Back in precalculus, we were taught that a function must only have one output per input. But, $\pm\sqrt{x}$ gives two outputs!

Edit
An answer to your edited problem. $y=x^2 \Rightarrow x=\pm\sqrt{y}$. On the left and the right side we take the square root. Don't forget the $\pm$!!
