Consider the following function: $$ F(x,y) \ = \ \frac{x^2y}{x^4+y^2} $$
Expanding $x=0$ and $y=0$, it is a cool example of a function whose series depends on the order in which you take the expansion: $$ \mathrm{about\ }x=0 \mathrm{\ first}\ \ \ \implies F(x,y) \approx\frac{x^2}{y} \\ \mathrm{about\ }y=0 \mathrm{\ first}\ \ \ \implies F(x,y) \approx\frac{y}{x^2} $$
There is clearly a very bad singularity at $(x,y) = (0,0)$ for this particular function.
My question is:
Does there exist a function $f(x,y)$ which for an expansion about some point $(x_0,y_0)$, there is a different series depending on the order in which you expand the variables where $f(x_0,y_0)$ is constant and not singular?
I am imagining if there is such a function, then the funny point $(x_0,y_0)$ would be a saddle point or something of that nature.
If I read this correctly, what you ask for is not a full expansion (Taylor) but the behavior of the function at $x=x_0$ and at $y=y_0$.
Consider
$$ F(x,y) = (1+x)(1+y)^2 $$ near $(x_0,y_0) = (0,0)$.
With your approach, at $x=0$ you have $(1+y)^2$, at $y=0$ you have $(1+x)$.
So this is different, however only at first sight. Because what this really shows is the behavior of $F(x,y)$ on the $y$-axis (for $x=0$) and on the $x$-axis (for $y=0$).
However, near $(x_0,y_0) = (0,0)$ the correct Taylor expansion to first order is $1 + x + 2y$ and this does not depend on the order of approximation. Multivariable Taylor expansion is guaranteed to work (which entails independence of any order of operations) if $F(x,y)$ and all (mixed in variables) partial derivatives are bounded and continuous at $(x_0,y_0)$.