Answer: At the point \(A(1,-2)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{4}{3}\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\dfrac{20}{27}\).

Differentiate implicitly:

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

giving

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

Now substitute \(x=1\) and \(y=-2\):

\(3(4)y'-\left(6(1)(-2)+3(1)^2y'\right)=0\)

\(12y'-(-12+3y')=0\)

\(12y'-3y'+12=0\)

\(9y'=-12\)

\(y'=-\frac{4}{3}.\)

Differentiate the first derivative equation again:

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

Substitute \(x=1\), \(y=-2\), \(y'=-\dfrac{4}{3}\):

\(12y''-12\cdot\frac{16}{9}-\left(6\cdot1\cdot\left(-\frac{4}{3}\right)+3y''+6\cdot1\cdot\left(-\frac{4}{3}\right)+6(-2)\right)=0.\)

So

\(12y''-\frac{64}{3}-(-8+3y''-8-12)=0\)

\(12y''-\frac{64}{3}-(3y''-28)=0\)

\(9y''+\frac{20}{3}=0\)

\(9y''=-\frac{20}{3}\)

\(y''=-\frac{20}{27}.\)