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.
Bijection $\mathbb{Z} \to \mathbb{N}$:
$$f(x) = \left|2x-\frac{1}{2}\right|+\frac{1}{2}$$
Injections $\mathbb{Z} \to \mathbb{N}$:
$$g(x) = f(2x)\quad \text{ or } \quad g'(x) = 2f(x)$$
Surjections $\mathbb{Z} \to \mathbb{N}$:
$$h(x) = f\left(\left\lfloor\frac{x}{2}\right\rfloor\right)
\quad \text{ or } \quad h'(x) = \left\lfloor\frac{f(x)}{2}\right\rfloor$$
I hope this helps $\ddot\smile$
Best Answer
For your first question, suppose that we had two elements of $A$ with the same coordinate: that is, we had $(a, b) \in A$ and $(a, c) \in A$. Then by definition of $A$, we have that $b^2 = a = c^2$, so $b^2 = c^2$. So does it follow that $b = c$? (It doesn't.)
For the second, you are correct that it is injective: For if $f(n) = f(m)$, then $$(2n, n + 3) = (2m, m + 3) \implies n + 3 = m + 3 \implies n = m$$ You're also correct that it's not onto, since it never hits $(1, 0)$. If it were onto, then you'd select an arbitrary element of $\Bbb{Z} \times \Bbb{Z}$ and find an element of $\Bbb{Z}$ mapping to it.