As the figure below shows, the graph of the implicit function $$x^x=y^y,(x,y >0)$$ composes of a straight line and an arc, which of the two have an intersection point $P$.
How to find the coordinates $(x_p,y_p)$ of $P$? Does there exist a closed-form solution?
On
Since $x^x=y^y$, hence ${\rm d}x^x={\rm d}y^y,$ i.e. $x^x(\ln x+1){\rm d}x=y^y(\ln y+1){\rm d}y.$ We obtain the derivative $$\frac{{\rm d}y}{{\rm d}x}=\frac{x^x(\ln x+1)}{y^y(\ln y+1)}=\frac{\ln x+1}{\ln y+1}.$$ Notice that $$x_p=y_p,$$ and $$\dfrac{{\rm d}y}{{\rm d}x}\bigg|_{x=x_p}=-1.$$ Thus, we can set up an equation as follows $$\frac{\ln x_p+1}{\ln x_p+1 }=-1.$$ Therefore,$$\ln x_p=-1.$$ As a result, $$x_p=\frac{1}{e}.$$ It follows that $$(x_p,y_p)=\left(\frac{1}{e},\frac{1}{e}\right).$$
Am I right？ Indeed，there exists a flaw here， because the expression of $\dfrac{{\rm d}y}{{\rm d}x}=\dfrac{\ln x+1}{\ln y+1}$ has no definition at $x=y=\dfrac{1}{e}$. How to remedy the problem?
Let's use the following theorem from multivariable calculus:
Taking the function $f=x^x-y^y$ we see that the gradient at the intersection point of yours must be zero (it is impossible for a nonzero planar vector to be perpendicular to two linearly independent vectors simultaneously).
Solving the system $$\nabla f=\left(x^x(1+\log(x)),y^y(1+\log(y)) \right)= \mathbf{0} $$ gives $$x=y=\mathrm{e}^{-1}. $$