If $k$ is an algebraically closed field of characteristic zero then the algebraic closure of $k((x))$ is $F = \cup k((x^{1/n}))$, the field of Puiseux series. In particular, the algebraic closure of $k(x)$ is contained in $F$. Now you want to look at the algebraic closure of $k(x,y)=k(x)(y)$, which is contained in the algebraic closure of $F(y)$, which in turn is contained in $\cup F((y^{1/n}))$. I think that gives you what you want.

If $k$ is an algebraically closed field of characteristic zero then the algebraic closure of $k((x))$ is $F = \cup k((x^{1/n}))$, the field of Puiseux series. In particular, the algebraic closure of $k(x)$ is contained in $F$. Now you want to look at the algebraic closure of $k(x,y)=k(x)(y)$, which is contained in the algebraic closure of $F(y)$, which in turn is contained in $\cup F((y^{1/n}))$. I think that gives you what you want.

Edit: An element of $\cup F((y^{1/n}))$ is a power series in a (fixed) root of $y$ all of whose coefficients are powers series in roots of $x$ but it's not clear that one can bound $m$ such that all coefficients are in $k((x^{1/m}))$ for a fixed $m$.