Let $$X = \{(z, w) \in \mathbb{C}^2 : z^2 + w^2 = 1\}.$$ Is there a holomorphic function $f : X \to \mathbb{C}$ which do not extend to a holomorphic function on all of $\mathbb{C}^2$?
I tried to find an example of the form $f(z, w) = \frac{1}{g(z, w)}$, where $g(z, w)$ is a holomorphic function which has zeros, but none that lie in $X$. But I couldn't find any.
On
Here is one reason that you could not find a counterexample. Say you have $\psi(z,w)$ a holomorphic function on $\mathbb{C}^2$ that has no zeroes on $\{\phi=0\}$, that is $\phi$ and $\psi$ have no common zeroes. Then take $\frac{1}{\psi(z,w)}$ and somehow get a contradiction. Now it turns out that if two $\phi$, $\psi$ have no common zeroes then there exist other two functions $a$, $b$ such that $$a \phi + b \psi = 1$$
Therefore, on $\{\phi=0\}$ we have $\frac{1}{\psi} = b$, which extends. Uff...
Now, a general approach to the extension of functions from $\{\phi=0\}$ to the whole space.
Let $f$ be holomorphic on $X=\{\phi=0\}$. There exists an open cover $(U_n)_n$ of the whole space $\mathbb{C}^2$ and a family of holomorphic maps $F_n$ on $U_n$ such that $$F_n\ _{\mid X\cap U_n} = f_{\mid X\cap U_n}$$ that is, $F_n$ locally extends $f$ from $X\cap U_n$ to $U_n$. Now, the problem is that $F_n$ do not piece together to form a function on $\mathbb{C}^2$. In other words,
$$F_m\ _{U_m \cap U_n} \not= F_n\ _{U_m \cap U_n}$$ just yet. However, $$F_m\ _{U_m \cap U_n\cap X} = F_n\ _{U_m \cap U_n\cap X}$$
Now, if for a holomorphic function $G$ on $U$ open subset of $\mathbb{C}^2$ we have $$G_{\mid U \cap \{\phi=0\}} = 0$$ then $$G = \phi\cdot H$$ for some holomorphic function on $U$ ( the converse is obvious). This is not difficult to prove, but requires an argument. Let's leave it for now.
In any case, we have
$$(F_m - F_n)_{\mid U_m \cap U_n} = \phi\cdot G_{m n}$$ for some function $G_{mn} \colon U_{mn} \to \mathbb{C}$, holomorphic.
Now, we would like to write $$G_{mn} = H_m- H_n$$ for some functions $H_\colon U_m \to \mathbb{C}$. That would give us $$F_m - F_n = \phi( H_m - H_n)$$ and so $$F_m - \phi \cdot H_m = F_n - \phi \cdot H_n$$ on $U_m\cap U_n$ and so $$\tilde F_n \colon = F_n - \phi \cdot H_n$$ piece together into a function on $\mathbb{C}^2$ that restricts to $f$.
How to get the $H_n$'s? We have $$G_{mn}+ G_{nk}+ G_{km}= 0$$ on $U_m\cap U_n\cap U_k$ so they form a cocycle (Cech cohomology). Now, every cocycle is in fact a coboundary ( this follows from some existence result for the $\bar \partial $ problem).
Recommended text: Hormander, Several Complex Variables
Note: this result holds in general: if $X$ is a complex submanifold of $\mathbb{C}^n$, and $f\colon X\to \mathbb{C}$ is holomorphic, then there exists $F\colon \mathbb{C}^n$ that restricts to $f$. The proof is similar, but there are complications. It would be essentially the same if $X$ was the zero set of $\phi$, with $0$ a regular value. In codimension larger than $1$ one deals with "sheaves of (ideals of) holomorphic functions", and some discipline is required.
Why is this more complicated than the smooth case? There we have the partition of $1$, something that does not exists in the holomorphic case.
Claim. Every holomorphic function on $X = \{\, (z,w) : z^2 + w^2 = 1 \,\} \subset \mathbb C^2$ extends to a holomorphic function on $\mathbb C^2$.
Proof. Introducing the new coordinates $$s = z + iw, \quad t = z - iw$$ we see that $$X = \{\, s \cdot t = 1 \,\}.$$ The projection $\mathbb C^2 \to \mathbb C, (s,t) \mapsto s$ induces an isomorphism $$X \stackrel{\cong}{\rightarrow} \mathbb C \setminus \{0\}.$$ By this isomorphism we can identify holomorphic functions on $X$ with holomorphic functions on $\mathbb C \setminus \{0\}$ in one coordinate $s$. We can already construct functions on $\mathbb C \setminus \{0\}$ that do not extend to $\mathbb C$ but still extend from $X$ to $\mathbb C^2$! For example $$f(s) = \frac{1}{s} = t. \tag{1}$$ Now we use this idea to prove our claim.
Let $f(s)$ be any holomorphic function on $\mathbb C \setminus \{0\}$. Since $f$ has (at most) an isolated singularity in $0$, we can write $f$ as a Laurent series $$f(s) = \sum_{k=-\infty}^\infty a_k s^k.$$ Since the only singularities of $f$ are in $0$ and in $\infty$, the two power series $$g(t) = \sum_{k=1}^\infty a_{-k} t^k, \quad h(s) = \sum_{k=0}^\infty a_k s^k$$ have infinite radius of convergence, so that $$f(s,t) = g(t) + h(s)$$ is an extension of $f$ to $\mathbb C^2$.