Choose a sequence $a_n$ of distinct points converging to $0$. "Choose" an ultrafilter on this countable set. Consider the germs of functions $f$ such that the set of all $a_n$ where $f(a_n)=0$ belongs to the filter. (This condition on the function depends only on its germ.)

EDIT The following refers to and complements George Elencwajg's answer. In particular the ideal $\frak m_a$ and $\frak m_a^{nb}$ are defined there.

Recall: An ultrafilter on the set $X$ gives you a maximal ideal in the ring of all real-valued functions, and these are the only prime ideals. If $X$ has a topology, then intersecting with $C(X)$ these yield prime ideals which may or may not be maximal. If $X$ is compact Hausdorff then every ultrafilter converges to (i.e. contains all neighborhoods of) a unique point.

For the rest of this discussion I assume $X$ is compact Hausdorff.

(1) Every proper ideal in $C(X)$ is contained in $\frak m_a$ for a (unique) point $a\in X$. In particular these are the only maximal ideals.

Proof: Otherwise $I$ has elements $f_1,\dots ,f_n$ such that the open sets $X-f^{-1}(0)$ cover $X$. But then $f_1^2+\dots +f_n^2$ is a a unit in $C(X)$.

(2) Every prime ideal $P$ of $C(X)$ must contain the ideal $\frak m_a^{nb}$ for some point $a\in X$.

Proof: Otherwise there exist $f_1,\dots ,f_n$ all outside $P$, vanishing respectively on open sets $U_1,\dots ,U_n$ that cover $X$. The product of the $f_i$ is zero, contradiction.

Thus

(3) All prime ideals of $C(X)$ correspond to prime ideals of the various local rings (rings of germs) $C(X)/\frak m_a^{nb}$.

Of course when the prime is determined by an ultrafilter then the point in question is the point that it converges to.

(4) The prime determined by an ultrafilter cannot be maximal unless the ultrafilter is principal.

Proof: If the ideal is $\frak m_a$, then a continuous function vanishing only at $a$ will belong to the ideal, implying that the singleton $a$ belongs to the filter.

I suppose it can happen that different ultrafilters sometimes give the same prime ideal, but I'm not sure.

I suppose it can happen that there are prime ideals not determined by any ultrafilter, but I'm not sure. Such an ideal would **not** have the following property : If $f\in I$ and if $f^{-1}(0)\subset g^{-1}(0)$ then $g\in I$.