Help with the graph of y^x=x^y on Cartesian coordinates While attempting to estimate the graph of equation, $y^x=x^y $ (domain is positive real numbers) I was able to deduce that  it contains two curves: a straight line $y=x$, and another curve (that LOOKS like a rectangular hyperbola but I'm pretty sure it's not) that is asymptotic at $x=1$ and $y=1$. Assuming it to be a rectangular hyperbola, and using its asymptotes and the fact that it cuts $y=x $   at  $ (e,e)$ , I came up with its equation as : $(y-1)(x-1)=(e-1)^2$  . At this moment I open up a graphing software (Desmos) to check my work and I find that the Hyperbola I came up with approximates the curve very well but is quite obviously not the right one...... My question now is, can we find an explicit equation for just that particular curve (excluding the $y=x$) ? And if not, how do I modify my hyperbolic approximation so that it gets accurate at a particular value of $x$ (for instance, in the above stated approximation, it gets very accurate at $x=e$ )
 A: Parametrisation:
The one type of graph is $(x;y):=(t;t)$ .
The other type can be discribed by $(x;y):=((1+\frac{1}{t})^t;(1+\frac{1}{t})^{t+1})$ .
A: Welcome to the world of Lambert function !
$$y^x=x^y \implies y=-\frac{x }{\log (x)}W\left(-\frac{\log (x)}{x}\right)$$ 
For $0 \leq x \leq e$, this is effectively a straight line $(y=x)$ and, for $x > e$, it "looks" like an hyperbola (but it is not).
The approximation you give is not bad at all; for the range $e \leq x \leq 15$, inspired by your model, I curve fitted
$$y=a+\frac b {(x-c)^{d} }$$ and obtained a quite good fit 
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 1.01812 & 0.00043 & \{1.01728,1.01896\} \\
 b & 2.44174 & 0.00358 & \{2.43464,2.44884\} \\
 c & 1.21741 & 0.00221 & \{1.21303,1.22178\} \\
 d & 0.89015 & 0.00113 & \{0.88792,0.89238\} \\
\end{array}$$
Edit
For a totally symmetric model, we can use, just by analogy with what you proposed 
$$(y^a-1)(x^a-1)=b \implies y=\left(1+\frac{b}{x^a-1}\right)^{\frac{1}{a}}$$  using your values as good initial estimates for the nonlinear regression, the curve fit for the same range as earlier gives as a very good fit
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 0.77050 & 0.00024 & \{0.77002,0.77097\} \\
 b & 1.34807 & 0.00124 & \{1.34561,1.35053\} \\
\end{array}$$ meaning that $a=\frac 34$ and $b=\frac 43$ would not be bad.
Repeating the calculations with
$$y=\left(1+\frac{1}{a(x^a-1)}\right)^{\frac{1}{a}}$$ leads again to a good fit
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 0.76320 & 0.00003 & \{0.76315,0.76326\} \\
\end{array}$$
