Explicit elements of $K((x))((y)) \setminus K((x,y))$

Suppose $R$ is a domain with field of fractions $F$. Let $f\in F[[y]]$ and suppose that $f\in Frac(R[[y]])$. Then $f=h/g$ with $g,h\in R[[y]]]$ and we may assume that $g=b_0+b_1y+\dots$ with $b_0\ne0$. Therefore $$ b_0g^{-1}=(1+(b_1/b_0)y+\dots)^{-1}\in R[[y/b_0]] $$ and so $f\in b_0^{-1}R[[y/b_0]]$.

So when $R=k[[x]]$, any $f=\sum y^n/x^{r(n)}$ with $r(n)/n \to \infty$ will work.

An analytic paraphrase of this argument (which is relevant to Makhalan's comment) is that (regarding $F=k((x))$ as a valued field) any element of $Frac(k[[x,y]])$ is a ratio of power series converging on the open unit disc in $F$. So it is enough to write down a power series in $y$ with zero radius of convergence.

There is a characterization of $K((x,y))$ which makes it clear what sorts of examples to expect. That is, it is a subring of $K(z)((y))$ (where $x = yz$) of series with certain regularity properties.

Let $R \subset K[z][[y]]$ be the "filtered power series" ring, consisting of series $p(z,y) = \sum_{n \geq 0} p_n(z) y^n$ with $\deg p_n \leq n$; equivalently, $p_n(x/y) y^n$ is homogeneous in $x,y$ of degree $n$. Then by definition the substitution $x = yz$ furnishes an isomorphism $R \cong K[[x,y]]$. One verifies that $R$ is indeed a ring and that if $p(z,0) = 1$, then $p$ is a unit in $R$, since $$p(z,y)^{-1} = (1 + \sum_{n \geq 1} p_n(z) y^n)^{-1} = \sum_{m \geq 0} (-1)^m (\sum_{n \geq 1} p_n(z) y^n)^m.$$

Now we consider $\mathrm{Frac}(R)$. If $p(z,y) = \sum_{n \geq a} p_n(z) y^n$ with $p_a \neq 0$, then we can write $$p(z,y) = p_a(z) y^a q(z,y)$$ in $K(z)[[y]]$, where if $q(z,y) = \sum_{n \geq 0} q_n(z) y^n$, then $q_n(x/y) y^n$ is homogeneous of degree $n$ (as a rational function). Then $p^{-1} = p_a^{-1} y^{-a} q^{-1} \in K(z)((y))$ also has this property.

Conversely, if $\ell(z,y) \in K(z)((y))$ is a "filtered Laurent series": $\ell(z,y) = \sum_{n \geq -a} \ell_n(z) y^n$ with $a \in \mathbb{Z}$, $\ell_n(z) \in K(z)$, and $\ell_n(x/y)y^n$ homogeneous of degree $n \in \mathbb{Z}$, then $\ell(x/y,y)$ makes sense in $K((x,y))$ if each $\ell_n$ is additionally of the form $q_{n + a}(z) p_a(z)^{-1}$ for polynomials $q_{n + a}, p_a$ of degrees at most $n + a, a$ respectively, since then $$\ell(x/y,y) = p_a(x/y)^{-1} y^{-a} \sum_{m \geq 0} q_m(x/y) y^m,$$ with $p_a(x/y)^{-1} \in K(x/y) \subset K((x,y))$ and $q_m(x/y) y^m \in K[x,y]$ homogeneous of degree $m$. More concisely, we can say: $K((x,y))$ is isomorphic to the subring of $K(z)((y))$ consisting of series $$\ell(z,y) = \sum_{n \geq -a} \ell_n(z) y^n$$ such that each $\ell_n(x/y)y^n$ is homogeneous of degree $n$ and all the $\ell_n(z)$ together have only finitely many poles in $\bar{K}$.

Thus, the following classes of functions in $K((x))((y))$ are not in $K((x,y))$:

• Series of the form $\ell(x,y) = t(x/y) = \sum_{n \geq 0} t_n x^n y^{-n}$ for which $t(z) \in K[[z]]$ is not a rational function. For example, $$t(z) = \exp(z) = \sum_{n \geq 0} \frac{1}{n!} z^n, \qquad t(z) = \sum_{n \geq 0} b_n z^n,$$ where the first only makes sense over a field of characteristic zero, and where in the second, $b_n \in \{0,1\}$ is a random sequence of bits (this will almost surely not be rational). More generally, series of the form $\ell(x,y) = \sum_{a,b} \ell_{a,b} x^a y^b$ such that some $\ell_n(z) = \sum_{a + b = n} \ell_{a,b} z^a$ is not rational.
• Series $\ell(x,y)$ for which the $\ell_n(z)$ have "infinitely many poles" among them. This is hard to detect on the level of formal series, but one criterion is similar to that of Tony Scholl: the radii of convergence of the $\ell_n(z)$ have infimum zero or infinity. Over $\mathbb{C}$, by the root test this is the same as $\lim_{n \to \infty} \sqrt[n]{|\ell_{a, n - a}|}$ being unbounded in $a$, or their inverses being unbounded. (This also works over any field with a complete real valuation.)

Addendum: actually there is a chain of subfields between $K((x,y))$ and $K((x))((y))$ with cardinality c. For instance, for any $\lambda>1$ we may consider the subset $R_\lambda$ of $K((x))((y))$ of all Laurent series $\sum_k c_k(x)y^k$ with $c_k\in K((x))$ satisfying $$\inf_k \lambda ^{-k} \mathrm{ord}(c_k) > -\infty.$$ It's easy to check that it's a subfield of $K((x))((y))$, containing $K[[x,y]]$.

Moreover, since in place of $\lambda^k$ we can use functions $\mathbb{N}\to\mathbb{N}$ with arbitrarily large growth, one can also show that the subfields between $K((x,y))$ and $K((x))((y))$ have uncountable cofinality also.