The equation of a curve is

\(y=x^2\sqrt{3+x}\)

for \(x\geq-3\).

(i) Find \(\dfrac{dy}{dx}\).

(ii) Find the equation of the tangent to the curve \(y=x^2\sqrt{3+x}\) at the point where \(x=1\).

(iii) Find the coordinates of the turning points of the curve \(y=x^2\sqrt{3+x}\).