Let $Y=\mathbb{A}^2-\{(0,0)\}$. Show that $K(Y)=k(x,y)$, and $\mathcal O(Y)=k[x,y]$. 
Let $Y=\mathbb{A}^2-\{(0,0)\}$.
Show that $K(Y)=k(x,y)$, and $\mathcal O(Y)=k[x,y]$.

where $K(Y)$ is the function field on $Y$, $\mathcal O(Y)$ is the set of regular functions.
I already know  if $U$ is nonempty open subset of $Y$ then $K(Y)=K(U)$ since $Y$ is irreducible.
Would it be useful for above question?
 A: The fact you mentioned is very useful for your question! But you need to think slightly differently.
You need to realise that $\mathbb A^2 - \{ (0,0) \}$ is itself an open subset of $\mathbb A^2$. So the function field of $\mathbb A^2 - \{ (0,0) \}$ is the same as the function field of $\mathbb A^2$, which is certainly $k(x,y)$.
There are many ways of doing the other part of the question. One method that comes to mind is to consider the open sets:
$$ U = \{ (x,y) \in \mathbb A^2 : x \neq 0 \}, \ \ \ \ \ \ V = \{ (x,y) \in \mathbb A^2 : y \neq 0 \}.$$
So $U \cup V = \mathbb A^2 - \{ (0,0) \} = Y$.
What is nice about $U$ and $V$ is that they are both complements of hypersurfaces in $\mathbb A^2$. Therefore, $\mathcal O(U)$ and $\mathcal O(V)$, the rings of regular functions on $U$ and $V$, can be determined using the Rabinowitsch trick. If you do this, the result is
$$ O(U) = k[x,y]_x, \ \ \ \ \ \ O(V) = k[x,y]_y.$$
Finally, a rational function is regular on $Y = \mathbb A^2 - \{ (0,0) \} = U \cup V$ iff it is regular on both $U$ and $V$. So
$$ O(Y) = k[x,y]_x \cap k[x,y]_y = k[x,y]$$
(Here I'm viewing both $k[x,y]_x$ and $k[x,y]_y$ as subrings of $k(x,y)$, and I'm taking the intersection inside $k(x,y)$.)
