"The only way I know how to solve this is implicit differentiation:"

Yeah but you took the natural log of both sides to simplify, so you did use logarithmic differentiation.

But man, you did something wierd. You took the derivative of ln y, and wrote 1/y * y' in accordance with the chain rule. I suppose this is the same as 1/y dy but you didn't do it on the other side. (dx)

Anyways, I'll attempt to do it using your method.

u = 2x

y = u^u

ln y = u ln u

1/y dy = u/u du + ln u du

1/y dy = du + ln u du

1/y dy = (1 + ln u) du

Now u = 2x so du = 2 dx

Inserting these values we get:

1/y dy = (1 + ln 2x)2 dx

dy/dx = y(1 + ln 2x) 2

Once again, I arrive at the same conclusion.