Is $(0,0)$ a solution to $x^y-y^x=0$? I am trying to determine if $(0,0)$ a solution to $x^y-y^x=0$.  My hunch is that it is undefined since $0^0$ is an indeterminate form.  To attempt to prove this, I have tried the usual "different paths give different limits" trick with
$(x,ax^n)\rightarrow(0,0)$
$(x,\sin(x))\rightarrow(0,0)$
$(x,e^x-1)\rightarrow(0,0)$
None of the above accomplished my goal.  
I did find this older post ($x^y = y^x$ for integers $x$ and $y$) which included an answer by "Yuval Filmus" which states $y=x=0$ is $\it{trivially}$ a solution.  He accomplished this by defining $0^0$ to some value and moves on.  I would like to see something more rigorous, if it exists.
Any hints on how to proceed?
Edited for wording.
Edit:  Stefan Smith confirms my suspicion that the comment of Yuval Filmus cannot be made rigorous.  
 A: This might sound unsatisfying, but if you define $0^0$ to be any finite number, then $(0,0)$ is a solution, otherwise it isn't. I personally like $0^0=1$, but some people disagree. This site has a lot to say about the issue of $0^0$, and you can search for that. It is ultimately a matter of definition. 
A: First of all, this function may not be well-defined if $x<0$ or $y<0$ (for example, if $x=-1$ and $y=1/2$, then $x^y=i$, which is not real). Therefore, I suggest you do the following:


*

*Define the function only on $\mathbb R_{++}^2\equiv\{(x,y)\in\mathbb R^2\,|\,x>0,\, y>0\}$

*Show that for any sequence $(x_n,y_n)_{n\in\mathbb Z_+}\subset\mathbb R_{++}^2$ that converges to $(0,0)$, you have that $f(x_n,y_n)$ converges to $0$.


In this way, you can continuously extend the function to $(0,0)$ by defining $f(0,0)\equiv0$.

Too bad this cannot be done! To see this, let
\begin{align*}
(x_n,y_n)\equiv\left(\frac 1 n,\frac{1}{\ln(n+1)}\right)\quad\forall n\in\mathbb Z_+
\end{align*}
Clearly, $(x_n,y_n)\to 0$ as $n\to\infty$. However,
\begin{align*}
f(x_n,y_n)=\left(\frac{1}{n}\right)^{1/\ln(n+1)}-\left(\frac{1}{\ln (n+1)}\right)^{1/n}
\end{align*}
converges to $1/e-1$ as $n\to\infty$. Because of this, $0$ is not a good definition of $f(0,0)$, since $0$ may not be approached by the function $f$ when evaluated close to $(0,0)$.

In terms of how you put it originally, try
$$\left(x,\ln\left(\dfrac{x+1}{x}\right)^{-1}\right)$$
This converges to $(0,0)$ as $x\searrow0$, but the function value converges to $1/e-1$.
A: The short answer is "no".  The reason is that there is no universal, always-agreed-upon definition of what $0^0$ is supposed to equal.  And if $0^0$ is undefined, so is $0^0-0^0$, and $(0,0)$ is not a solution of $x^y-y^x=0$.
Having said that, there are many situations in which it is convenient to allow $0^0$ to take a value, as long as you are extremely careful, you have specified what number system you are using (reals? integers? nonnegative integers? complex numbers?) and what rules you are operating under.  In most such situations I have seen, that value is $1$.
Your question does not provide any such context.  There is a (diophantine-equations) tag on your question, but your question does not mention Diophantine equations, and whether $x$ and $y$ are integers.  So one can't really make a good case that $(0,0)$ solves your equation.  
