This is incorrect. x and y are both functions of t, so when you differentiate both sides, you should get
$\displaystyle \displaystyle \begin{align*} x^2 + y^2 &= 100 \\ \frac{d}{dt} \left( x^2 + y^2 \right) &= \frac{d}{dt} \left( 100 \right) \\ \frac{d}{dx}\left( x^2 \right) \frac{dx}{dt} + \frac{d}{dy}\left( y^2 \right) \frac{dy}{dt} &= 0 \\ 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} &= 0 \\ 2x\,\frac{dx}{dt} + 2 \, \sqrt{ 100 - x^2 } \,\frac{dy}{dt} &= 0 \end{align*}$
NOW you can substitute in the known values and solve for $\displaystyle \displaystyle \frac{dx}{dt}$.