morphism from a local ring of a scheme to the scheme

Let me add to Amitesh's absolutely correct answer a few words describing the image $I\stackrel {\text {def}}{=}Im(f)$ of the canonical morphism $f:\text{Spec} (\mathcal O_{X,x}) \to X$ .

a) The set $I$ is exactly the intersection of all neighbourhoodz of $x$ in $X$: it is a kind of microgerm of $X$ at $x$.
Beware that $I$ is not a subscheme of $X$ since it is not locally closed.

b) More geometrically (and thus more interestingly!) consider the irreducible subvariety $V=\overline {\lbrace x\rbrace}\subset X$ whose generic point is $x$.
Let $Y\subset X$ be a closed irreducible subscheme on which $V$ lies: $V\subset Y$ and let $\eta_Y$ be the generic point of $Y$.
Then our subset $I$ is exactly the set of all those generic points $\eta_Y$. We say that $I$ is the set of generizations of $x$.

c) Two examples:
1) If $X$ is an irreducible scheme with generic point $\eta$, then for $x=\eta$ we have $I={\lbrace \eta\rbrace}$.
2) If $X=\mathbb A^2_\mathbb C=\text {Spec}(\mathbb C[x,y])$ and $x=(a,b)$ (more accurately $x$ is the maximal ideal $\mathfrak m= (x-a,x-b)$) , then $I$ is the set consisting in $x$, the generic point of $\mathbb A^2_\mathbb C$ and the generic points of all irreducible curves going through $x$, like for example the curve $(y-b)^2-(x-a)^3=0$.

(1) Let $U$ and $V$ be open affine subsets of the scheme $X$ such that $x\in U\cap V$. Choose an open affine subset $x\in W\subseteq U\cap V$. Prove that the compositions $\text{Spec}(O_{X,x})\to W\to U\to U\cup V$ and $\text{Spec}(O_{X,x})\to W\to V\to U\cup V$ are equal to the composition $\text{Spec}(O_{X,x})\to W\to U\cap V\to U\cup V$. (Hint: recall that $U,V,W$ are affine open subsets of $X$ and you understand affine schemes by commutative algebra!)

(2) If $A$ is a commutative ring and if $p$ is a prime ideal, then the spectrum of the localization homomorphism $A\to A_{p}$ is the map $\text{Spec}(A_p)\to \text{Spec}(A)$. The image of this map equals the set of all prime ideals of $A$ contained in $p$ (prove this fact from commutative algebra if it is not obvious).

Here's a point of view that combines aspects of both of the current (good) answers but makes things more explicit.

Take $U$ an affine open piece of $X$ containing $x$, $U \cong \operatorname{Spec}(A)$, with $x$ corresponding to $P < A$. Then the localisation map $\phi:A\rightarrow A_P$ induces a morphism $\operatorname{Spec}(\mathcal{O}_{X,x})\rightarrow \operatorname{Spec}(A)\rightarrow X$. The image of this morphism is the set of primes in $A$ that are themselves contained in $P$. (Using properties of localisation of rings of morphisms between affine schemes.) This is equal to the intersection of every open subset of $\operatorname{Spec}(A)$ containing $P$, as is easily verified by checking for distinguished affine open subsets of $U$, $D(f)$ for $f \in A\setminus P$.

Now suppose $f \in A/ P$. Then the image of $\phi$ lies inside the distinguished open affine piece $D(f)$. Equivalently, $\phi$ factors through $A_f$, i.e. there is $\phi_f:A_f\rightarrow A_P$, such that if $i_f:A\rightarrow A_f$ is the localisation map, $\phi = \phi_f\circ i_f$. This means that the morphism $\operatorname{Spec}(\mathcal{O}_{X,x})\rightarrow\operatorname{Spec}(A_f)\rightarrow X$ induced by taking affine open piece $U' \cong \operatorname{Spec}(A_f)$ of $X$ is equal to the morphism induced by $U$. (Here we are implicitly using that for a manipulatively closed subset $S$ of a ring $R$, with $P$ a prime not meeting $S$, $(S^{-1}R)_{S^{-1}P}$ is naturally isomorphic to $R_P$.)

Thus if $V \cong \operatorname{Spec}(B)$ is another open affine piece of $X$ containing $x$, then taking open $W \subset U\cap V$ a distinguished open affine piece of both $U$ and $V$ (i.e. there is $a \in A, b \in B$ such that $W \cong D(a)$ and $W \cong D(b)$), the morphism $\operatorname{Spec}(\mathcal{O}_{X,x})\rightarrow X$ induced by $A$ is equal to the morphism induced by $A_a$ by the above paragraph, which is equal to the morphism induced by $B_b$ from the construction of $W$, which is then equal to the morphism induced by $B$ using the above paragraph again.

So we have shown that the morphism is independent of choice of open affine piece. Given this, the image is immediately seen to be the intersection of every open set about $x$, and (interestingly) this view is completely independent of the choice of affine piece.