All real values of '$u$' such that the curves $y=x^2+u$ and $y=\sqrt{x-u}$ intersect 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, ${\left(-1\right)}^2+4\left(1\right)\left(u\right)=1+4u$ , 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 $u\left(u+1\right)<0$, or $u\in \left(-1,0\right)$.