All real values of $\text{'}u\text{'}$ such that the curves $y=x^2+u$  and $y=\sqrt{x-u}$  meet in exactly one point is

There are two possibilities : either the curves  $y=x^2+u$ and  $x=y^2+u$ intersect in exactly one point, or they intersect in two points but one of the points occurs on the branch $y=-\sqrt{x-u}$ .
Case-1 : The two curves are symmetric about y = x, so they must touch that line at exactly one point and not cross it. Therefore, $x=x^2+u$ , so $x^2-x+u=0$ . This has exactly one solution if the discriminant,${(-1)}^2+4\left(1\right)\left(u\right)=1+4u$ , equals 0,so  $u=\boxed{\frac{1}{4}}$.
Case-2 :  $y=x^2+u$ intersects the x–axis at $\pm \sqrt{-u}$ , while $y=\sqrt{x-u}$ starts x = u and goes up from there. In order for these to intersect in exactly one point, we must have $-\sqrt{-u}<u$ , or  $-u>u^2$ (note that $-u$  must be positive in order for any intersection points of $y=x^2+u$  and  $x=y^2+u$ to occur outside the first quadrant). Hence we have $u(u+1)<0$ , or  $u\in (-1,0)$.