Taylor expansion of $f(x, y)$ when $x/y<<1$, but there are lonely $x$'s and $y$'s.

Question

I want to do a Taylor expansion of a specific function $f(x, y)$ when $x/y<<1$. The problem is that although the function has the fraction $x/y$, it also has lonely $x$'s and $y$'s that cannot be expressed as the fraction $x/y$. To give an easy example function, consider \begin{equation} f(x, y) = \sqrt{x + \frac{x}{y}}. \end{equation} Question 1: Does the Taylor expansion exist?
Question 2: If it does, how does one calculate it?

Answer 1

I looked into the problem and I'm going to post an answer myself. I'm not sure if this is correct so it would be helpful if someone could confirm that I'm right.

The answer: You are supposed to make a new variable $u=x/y$. Then you expand the function at $u=0$ and in the end go back to $u=x/y$. For the example function above, you would write \begin{equation} f(x, \frac{x}{u}) = \sqrt{x + u} \end{equation} and then the Taylor expansion to the leading order at $u=0$ would be \begin{equation} f(x, \frac{x}{u}) = \sqrt{x} + \frac{u}{2\sqrt{x}}. \end{equation} Finally, substitute $u=x/y$. \begin{equation} f(x, y) = \sqrt{x} + \frac{\sqrt{x}}{2y} = \sqrt{x}\left(1 + \frac{1}{2y}\right) \end{equation}