When two curves are tangent, they share the same line of tangent, i.e. the at slope at the same point. The derivative of the first curve is $y' = 2x$ and the derivative of the second curve is $y' =\frac{1}{2y}=\frac{1}{2\sqrt{x}}$. Setting them equal gives $$2x = \frac{1}{2\sqrt{x}}\implies x=2^{-\frac{4}{3}}$$

Setting this value to $y^2=x$ finds the tangent point is $\left(2^{-\frac{4}{3}}, 2^{-\frac{2}{3}}\right)$. Therefore, $$c=y-x^2=2^{-\frac{2}{3}}-\left(2^{-\frac{4}{3}}\right)^2=\boxed{\frac{3}{4\sqrt[3]{4}}}$$