Answer: (i) The point is \(A=(2,-2)\).

(ii) The value of \(\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}\) at \(A\) is \(\frac{1}{2}\).

Differentiate the curve implicitly:

\(2x^3+3x^2y-3y^3-16=0\)

gives

\(6x^2+6xy+3x^2\frac{\mathrm{d}y}{\mathrm{d}x}-9y^2\frac{\mathrm{d}y}{\mathrm{d}x}=0\).

So

\(6x^2+6xy=(9y^2-3x^2)\frac{\mathrm{d}y}{\mathrm{d}x}\),

hence

\(2x(x+y)=(3y^2-x^2)\frac{\mathrm{d}y}{\mathrm{d}x}\).

(i) At the point where \(\frac{\mathrm{d}y}{\mathrm{d}x}=0\) and \(x\neq 0\), the left-hand side must be zero, so \(x+y=0\). Thus \(y=-x\).

Substitute into the curve:

\(2x^3+3x^2(-x)-3(-x)^3-16=0\)

\(2x^3-3x^3+3x^3-16=0\)

\(2x^3=16\), so \(x=2\). Then \(y=-2\).

Therefore \(A=(2,-2)\).

(ii) Differentiate \(6x^2+6xy+3x^2y'-9y^2y'=0\) again with respect to \(x\):

\(12x+6(y+xy')+6xy'+3x^2y''-\left(18yy'^2+9y^2y''\right)=0\).

At \(A=(2,-2)\), we have \(y'=0\), so this becomes

\(12(2)+6(-2)+3(2^2)y''-9(-2)^2y''=0\).

So

\(24-12+12y''-36y''=0\)

\(12-24y''=0\)

\(y''=\frac{1}{2}\).