Note that $y=x^2$ is an equation, to be more precisely a statement form. It states: “The second coordinate is the square of the first one.” For the coordinates of a point that statement may be true -- for $(-2,4)$, e.g.-- or false for, say $(7,50)$. The set of points for which the statement form yields a true statement is called the solution set of the statement form. The statement form $y-x^2=0$ states: “The difference of the first coordinate and the square of the second equals zero.” is a different statement form, but has the same solution set as the first one.
The second one is given in a so called implicit form, wheras the first is called an explicit form. Why bother about the to forms? Well, if you want to determine the second coordinate of a point in the solution set, given the first coordinate is $7$, let's do it using the imlicit form. We have
$$y-7^2=0\iff y=49,$$
so you have to solve an equation. (That's the reason for why we call it implicit.). Using the explicit form we get immediately $y=7^2=49$ without solving any equation.
Your first example $y=1/x$: “The second coordinate is the reciprocal of the first one.” reads in implicit version $xy=1$: “The product of both coordinates equals one.” Another explicit form is $x=1/y$. So if you are able to isolate one coordinate in an implicit form, this coordinate is called dependent, the other independent. In $x=1/y$ the coordinate $x$ is dependent and $y$ independent whereas in $y=1/y$ it's vice versa.
Now in $y-x^2=0$ you can't isolate $x$ in a closed form, here we have $\sqrt{y}=|x|$. So if for example $y=25$ we have $\sqrt{25}=|x|\iff5=|x|\iff x=5\lor x=-5$, so we have to solve an equation anyway. You may write $x=\pm\sqrt{y}$ and see that for all positive $y$ you'll get two values of $x$, so that explicit form is not a function, which mathematicians prefer.
In the example $x^2+y^2=1$ given by James there doesn't exist any explicit form which is a function, for an obvious reason: the solution set is the unit circle, centered at the origin.
Facit: if a statement form allows an explicit form of one of the coordinates and that explicit form is a function, you are free to call the isolated coordinate dependent and the other independent. I never use those attributes, because they are rather useless, won't gain any knowledge and are causing a lot of trouble as we see here.
suppose $x=n+p\\0\leq p<1$
$$\quad{f(x)=\sqrt{4[x]-[4x]}=\\\sqrt{4[n+p]-[4(n+p)]}=\\
\sqrt{4n-[4n+4p]}=\\
\sqrt{4n-4n-[4p]}=\\
\sqrt{-[4p]}\\\implies-[4p]\geq 0\\ [4p]\leq 0 \implies [4p]=0 \\0 \leq 4p<1 \\0\leq p <\frac{1}{4} \to\\D_{f(x)}=[n,n+\frac{1}{4}) ,n \in \mathbb{Z}\\R_{f(x)}=\{\sqrt{0}\}=\{0\} }$$
Best Answer
Your explanation is correct, but try to be more rigorous in your answers. For example, for the domain: $$y^2 >0 \implies x > -6 \Leftrightarrow x \in \left[-6,\infty\right).$$
Another way to see that $y$ does not define a function, is by indentifying the equation of parabola.
More specifically, The pencil of conic sections with the $x$ axis as axis of symmetry, one vertex at the origin $(0, 0)$ and the same semi-latus rectum $p$ can be represented by the equation:
$$y^2 = 2px + (e^2-1)x^2, \quad e \geq 0.$$
By your expression, the eccentricity is $e=1$ and $p=1/2$. Then, you can just shift this parabola by $+6$ and it maintains the same properties (meaning that $x$ will be the axis of symmetry).
Being able to plot something in cartesian coordinates doesn't mean that it represent a function, as you saw yourself. The definition of a function is strict and that answers your comment question "what's wrong with it".