If $M=(x,y)\subset k[X,Y]/(Y^2-X^3)=k[x,y]\stackrel{\text {def}}{=}A \;$, the maximal ideal you are interested in is $\mathfrak m=MA_M \subset A_M$ the maximal ideal of the local ring $A_M=\mathcal O_{(0,0),W}$.
The $k$-vector space $\mathfrak m/\mathfrak m^2$ is generated by $\bar x$ and $\bar y$ and all you have to check is that they are linearly independent.
A linear dependence relation $q\bar x+r \bar y=0$ ($q,r\in k$) means $qx+ry \in \mathfrak m^2=(x^2,xy,y^2)$.
Lifting to actual polynomials this means $$qX+rY\in (X^2,XY,Y^2)+(Y^2-X^3)=(X^2,XY,Y^2)\subset k[X,Y]$$ and implies $q=r=0$.
So $\bar x$ and $\bar y$ are linearly independent in $\mathfrak m/\mathfrak m^2$ and $dim_k(\mathfrak m/\mathfrak m^2)=2$
If the map from $X$ to the maxspec of $\mathcal O(X)$ is a bijection, then $X$ is indeed affine.
Here is an argument:
By assumption $X \to $ maxspec $\mathcal O(X)$ is bijective, thus quasi-finite,
and so by (Grothendieck's form of) Zariski's main theorem, this map factors as an open embedding of $X$ into a variety that is finite over maxspec $\mathcal O(X)$.
Any variety finite over an affine variety is again affine, and hence $X$ is an open subset of an affine variety, i.e. quasi-affine. So we are reduced to considering the case when $X$ is quasi-affine, which is well-known and straightforward.
(I'm not sure that the full strength of ZMT is needed, but it is a natural tool
to exploit to get mileage out of the assumption of a morphism having finite fibres, which is what your bijectivity hypothesis gives.)
In fact, the argument shows something stronger: suppose that we just assume
that the morphism $X \to $ maxspec $\mathcal O(X)$ has finite non-empty fibres,
i.e. is quasi-finite and surjective.
Then the same argument with ZMT shows that $X$ is quasi-affine. But it is standard that the map $X \to $ maxspec $\mathcal O(X)$ is an open immersion when $X$ is quasi-affine,
and since by assumption it is surjecive, it is an isomorphism.
Note that if we omit one of the hypotheses of surjectivity or quasi-finiteness, we can find a non-affine $X$ satisfying the other hypothesis.
E.g. if $X = \mathbb A^2 \setminus \{0\}$ (the basic example of a quasi-affine,
but non-affine, variety), then maxspec $\mathcal O(X) = \mathbb A^2$, and the open immersion $X \to \mathbb A^2$ is evidently not surjective.
E.g. if $X = \mathbb A^2$ blown up at $0$, then maxspec $\mathcal O(X) =
\mathbb A^2$, and $X \to \mathbb A^2$ is surjective, but has an infinite fibre
over $0$.
Caveat/correction: I should add the following caveat, namely that it is not always true, for a variety $X$ over a field $k$, that $\mathcal O(X)$ is finitely generated over $k$, in which case maxspec may not be such a good construction to apply, and the above argument may not go through. So in order to conclude that $X$ is affine, one should first insist that $\mathcal O(X)$ is finitely generated over $k$, and then that futhermore the natural map $X \to $ maxspec $\mathcal O(X)$ is quasi-finite and surjective.
(Of course, one could work more generally with arbitrary schemes and Spec rather than
maxspec, but I haven't thought about this general setting: in particular, ZMT requires some finiteness hypotheses, and I haven't thought about what conditions might guarantee that the map $X \to $ Spec $\mathcal O(X)$ satisfies them.)
Incidentally, for an example of a quasi-projective variety with non-finitely generated ring of regular functions, see this note of Ravi Vakil's
Best Answer
We have an onto homomorphism $\mu:\mathcal{O}_X(U)\longrightarrow k$ defined by $\mu(f)=f(p)$ for all $f\in \mathcal{O}_X(U)$. Then $\ker(\mu)=I_U(p)$ and hence it is maximal.