(b)
The equation of given circle is
$x^{2} + y^{2} = 1$
On differentaing w.r.t. $t$, we get
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
But we have, $x = \frac{1}{2},y = \frac{\sqrt{3}}{2}$ and $\frac{dy}{dt} = - 3$, then
$\frac{1}{2}\frac{dx}{dt} + \frac{\sqrt{3}}{2}( - 3) = 0\ \ \ \Rightarrow \ \ \ \frac{dx}{dt} = 3\sqrt{3}$