Continuous function implicitly defining a non-continuous function (update) Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ be continuous on $(0,0)$ implicitly defining a function $y=y(x)$ on $\mathbb{R}$ (satisfies the hypothesis for the global implicit functions, or Dini's, theorem). Also $f(0,0)=0$.
I know for sure that is possible for the function $y(x)$ not to be continuous at the origin, but I couldn't find an example for such function.
Note: The global implicit functions theorem states:
Let $f:[a,b]\times[c,d]\rightarrow\mathbb{R}$ be continuous on the domain; if
1) $f(x,c)\cdot f(x,d)< 0~ ~~~~\forall x\in[a,b]$.
2) fixed $x_0\in[a,b]$, $f(x_0, \cdot)$ is strictly monotonic in $[c,d]$.
Then $\exists!~~y(x)$ such that $f(x,y(x))=0$ for each $x\in[a,b]$

I don't know if this answers your question, but let's try.
Consider $g:\mathbb{R}\to\mathbb{R}$ as
$$g(x)=\begin{cases}1 && \mbox{if }x\not=0\\0 &&\mbox{if }x=0\end{cases}$$
which is clearly not continuos in $x=0$. Now let $f:\mathbb{R}^2\to\mathbb{R}$ be
$$f(x,y)=\begin{cases}x^2+y^2 && \mbox{if }y\not=1\ \vee\ x=0\\0 &&\mbox{if }y=1\ \wedge\ x\not=0\end{cases}$$
Of course $f$ is continuos in $(0,0)$ - it's continuos almost everywhere actually - and, without using any theorem, you can see that $f(x,y)=0$ implicitly defines a function $y=y(x)$ such that $f(x,y(x))=0$. And, obviously, $y(x)=g(x)$.
