don’t

Yeah, one can perfectly get a similar answer without a calculator, too. First, try something in between 2 and 3 like 5/2, you get f(5/2) = (3125/32)^(1/2) ~= 10. Now take the second term of the Taylor expansion (or which is the same, do one iteration of the Newton method) at y = 5/2. We get f'(y)(x-y) = -1, so an approximate solution x would be x = y - 1/f'(y) = y - 1/(y^y(ln y + 1)) ~= 5/2 - 1/(10*(ln(5/2)+1)) ~=5/2 - 1/(10*(1+1)) = 5/2 - 1/20 = 2.45.