A very interesting question: intersection point of $x^y=y^x$
A Simple Approach
The simplest approach that I have found is to look at the intersections of
$$
y=tx\qquad\text{and}\qquad x^y=y^x\tag1
$$
That is,
$$
\begin{align}
x^{tx}&=(tx)^x\tag{2a}\\[3pt]
x^t&=tx\tag{2b}\\[3pt]
x^{t-1}&=t\tag{2c}\\
x&=t^{\frac1{t-1}}\tag{2d}\\
y&=t^{\frac t{t-1}}\tag{2e}
\end{align}
$$
Explanation:
$\text{(2a)}$: $x^y=y^x$
$\text{(2b)}$: raise to the $\frac1x$ power
$\text{(2c)}$: divide by $x$
$\text{(2d)}$: raise to the $\frac1{t-1}$ power
$\text{(2e)}$: $y=tx$
Now, if we wish to find where $x=y$, let $t\to1$. That is,
$$
\begin{align}
x
&=\lim_{t\to1}t^{\frac1{t-1}}\tag{3a}\\
&=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{3b}\\[6pt]
&=e\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: $x=y$ when $t=1$
$\text{(3b)}$: $t=1+\frac1n$
$\text{(3c)}$: evaluate the limit
Further Musings
We can also compute
$$
\begin{align}
y
&=\lim_{t\to1}t^{\frac t{t-1}}\tag{4a}\\
&=\lim_{n\to\infty}\left(1+\frac1n\right)^{n+1}\tag{4b}\\[6pt]
&=e\tag{4c}
\end{align}
$$
Explanation:
$\text{(4a)}$: $x=y$ when $t=1$
$\text{(4b)}$: $t=1+\frac1n$
$\text{(4c)}$: evaluate the limit
Using the results from this answer, we see that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing, which means that $\left(1+\frac1n\right)^n\le e\le\left(1+\frac1n\right)^{n+1}$.
Thus, as $t\to1^+$, $(4)$ and $(5)$ show that $x\to e^-$ and $y\to e^+$.
Furthermore, if we substitute $t\mapsto1/t$, we get $$ \begin{align} t^{\frac1{t-1}} &\mapsto(1/t)^{\frac1{1/t-1}}\tag{5a}\\ &=t^{\frac t{t-1}}\tag{5b} \end{align} $$ and $$ \begin{align} t^{\frac t{t-1}} &\mapsto(1/t)^{\frac{1/t}{1/t-1}}\tag{6a}\\ &=t^{\frac1{t-1}}\tag{6b} \end{align} $$ That is, substituting $t\mapsto1/t$ swaps $x$ and $y$.
This means that, as $t\to1^-$, $x\to e^+$ and $y\to e^-$.
Relation to the Graph
Here is where these points sit on the graph as $t\to1^+$:
Not sure to understand well the question but if it provides any help, I think there is something missing in your analysis.
Your equation can be seen as $g(x,y) = x^{y} = y^{x} = f(x,y)$. You are looking for intersecting points or surfaces (x,y) in $\mathbb{R}^{3}$, since both $f(\cdot,\cdot)$ and $g(\cdot,\cdot)$ are 2D scalar fields. Otherwise, if you are thinking the expression $x^{y} = y^{x}$ as the equation of a curve in $\mathbb{R}$ (somehow you figure out to express $y = f(x)$ or $x = f(x)$, it doesn't make sense to talk about an intercept (or interception set) since you must provide another curve or surface.
The same applies with the fields $g$ and $f$. You must provide another equation since you have two unknowns.
The answer you got $(x,y) = (e,e)$ is the trivial answer, and works for any real $k \in \mathbb{R}$: $(x,y) = (k,k)$.