What is 0 to the 0 is it one

lim[y→0-] (lim[x→0+] x^y) = lim[y→0+] (lim[x→0+] x^(-y))
= lim[y→0+] (lim[x→0+] 1/(x^y)) = 1/0 = ∞

If you 'fly' in along the line y=x, z goes to 1:
lim[x→0+] x^x = lim[x→0+] e^(x lnx) = e^(lim[x→0+] x lnx)
= e^(lim[x→0+] lnx /(1/x)) ... ~(∞/∞), so use l'Hôpital's Rule
= e^(lim[x→0+] [1/x] /[-1/x²])
= e^(lim[x→0+] [-x])
= e^0 = 1

If you 'fly' in along the line y=-x, z still goes to 1.
(Just re-work the y=x flight path, but starting with x^(-x).)

And you can choose various curved paths into the origin, that can produce other results, if the curve bends toward tangency with the y-axis at the origin.