
For $y=0$ we have $z=x^2$ and, therefore, along the curve
$$ ds^2 = df^2+dx^2 =(2x\,dx)^2+dx^2 = (4x^2+1)\,dx^2\,.$$
With $ds'=\sqrt{4x'^2+1}\,dx'$ we integrate then
$$ s = \int_0^s ds' = \int_0^x\sqrt{4x'^2+1}\,dx' = \frac{x}{2}\,
   \sqrt{4x^1+1} + \frac{1}{4}\,\ln\left(\sqrt{4x^2+1}+2x\right)\,,$$
where the integration was done with the online integral calculator
from Mathematica. By bisection of the function $s(x)$ we obtain for 
$s=10$ the value 
$$ x=3.041\,. $$

