Does the function $f(x)=x^{1/x}$ have any practical applications? I know that the Lambert function helps with $f(x) = x^{x}$ and there are related applications ($n^{n-2}$ is the number of spanning trees of the complete graph $K_{n}$). How about with the reciprocal in the exponent?
 A: $x^{\frac{1}{x}}$ actually comes up in solving the equation $x^y=y^x$. This can be rewritten as
$$
x^{\frac{1}{x}}=y^{\frac{1}{y}}
$$
by taking the $x$ and $y$ roots on both side, essentially dividing the exponents. If you are more comfortable with logarithms, the same property can be represented through
$$
yln(x)=xln(y)
$$
$$
\frac{1}{x}ln(x)=\frac{1}{y}ln(y)
$$
Regardless, from here you can solve for either $x$ or $y$ as a result of some clever algebraic manipulation:
$$
\frac{1}{x}ln(x)=\frac{1}{y}ln((y^{-1})^{-1})
$$
$$
\frac{1}{x}ln(x)=-\frac{1}{y}ln(y^{-1})
$$
$$
\frac{1}{x}ln(x)=-y^{-1}ln(y^{-1})
$$
$$
-\frac{1}{x}ln(x)=y^{-1}ln(y^{-1})
$$
$$
W\left(-\frac{1}{x}ln(x)\right)=W\biggl(y^{-1}ln(y^{-1})\biggr)
$$
$$
W\left(-\frac{1}{x}ln(x)\right)=ln(y^{-1})
$$
$$
W\left(-\frac{1}{x}ln(x)\right)=-ln(y)
$$
$$
-W\left(-\frac{1}{x}ln(x)\right)=ln(y)
$$
$$
y=e^{-W\left(-\frac{1}{x}ln(x)\right)}=\frac{W\left(-\frac{1}{x}ln(x)\right)}{-\frac{1}{x}ln(x)}
$$
I personally think this is just such a cool way to solve this equation, though you still have to approximate any answers you get except for very special values of $x$
