Analysis of the function $y=x^{\frac{1}{x}}$ The graph of the function $y=x^{\frac{1}{x}}$ for positive $x$ is as shown below:

When I calculated $y$ for negative values of $x$ only some of the values between $0$ and $-1$ and only those for which $x$ is odd(whole number), were given by Excel, and they are:

If the function behaves so nicely for positive $x$, why not so nicely for negative $x$?
If $y=-x^{\frac{1}{-x}}=\frac{1}{-x^{\frac{1}{x}}}$, why don't we take the reciprocal of all the values of $y$ for positive $x$ and negate them to plot the negative X-axis?
 A: The complex $\log$ map is multivalued and this forces $f$ to be multivalued for $x<0$.
$$\log(k,z)=\ln|z|+(\arg(z)+2k\pi)i,\,\,\,k\in\mathbb{Z}$$
If $z<0$, then $\arg(z)=\pi$
and therefore,
$$[z^{1/z},k]\to\exp\left(\frac{\log(k,z)}{z}\right),\,\,\,k\in\mathbb{Z}$$
consequently for negative $z$ the function is equivalent to the multivalued map:
$$[z^{1/z},k]\to\exp\left(\frac{\ln|z|+(\pi+2k\pi)i}{z}\right),\,\,\,k\in\mathbb{Z}$$
Some Maple code for verification:
 restart;
 f := proc (x) options operator, arrow; x^(1/x) end proc
 fn := proc (k, x) options operator, arrow;
 exp((ln(abs(x))+I*(Pi+2*k*Pi))/x) end proc

and now check:
 f(-2/3); evalc(%)

(3/4*I)*sqrt(2)*sqrt(3)
 fn(0, -2/3); evalc(%) #check using principal branch of log

(3/4*I)*sqrt(6)
A: For negative values of $x$, Excel seems to give you an answer for $x^{1/x}$ only when it knows the exponent $1/x$ is a rational number that can be written with an odd denominator.
For example, $(-3)^{-1/3} = 1/\sqrt[3]{-3}$.
If it is doing this, then you probably won't always have a negative answer. For example, for $x = -3/2$, you'd expect to get $1/(\sqrt[3]{-3/2})^2$, which is positive. 
