Answer: (i) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=1\).

(ii) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\dfrac{4}{3}\).

Let \(p=\dfrac{\mathrm{d}y}{\mathrm{d}x}\). At \(x=0\), \(y=-1\), so the equation gives

\((0+p)^3=(-1)^2+0\),

hence \(p^3=1\) and therefore

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=1\).

Now differentiate

\(\left(x+\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^3=y^2+x\)

with respect to \(x\):

\(3\left(x+\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\left(1+\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)=2y\dfrac{\mathrm{d}y}{\mathrm{d}x}+1.\)

Substitute \(x=0\), \(y=-1\), and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=1\):

\(3(1)^2\left(1+\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)=2(-1)(1)+1=-1.\)

So

\(1+\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\dfrac13\),

and therefore

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\dfrac43\).