i got 5/sqrt(3)
Taking the derivatives of the pythagorean theorem,
$x\left(\frac {dx}{dt}\right) + y\left(\frac {dy}{dt}\right) = z\left(\frac {dz}{dt}\right)$
But $\left(\frac {dz}{dt}\right) = 0$ so
$x\left(\frac {dx}{dt}\right) = - y\left(\frac {dy}{dt}\right)$
Then from pythagoras, $x^{2} + y^{2} = z^{2}$. We are looking for $x$ at $y = 10$ so $x = \sqrt {400 - 100} = \sqrt {300}$.
Then we have
$\sqrt {300}\left(\frac {dx}{dt}\right) = - 10(5)\Rightarrow\left(\frac {dx}{dt}\right) = \frac {50}{\sqrt {300}} = \frac {5}{\sqrt {3}}$
The negative sign just gives direction so it isn't really important.