Answer: (a) At the point \((-2,-1)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{1}{2}\).

(b) At the point \((-2,-1)\), \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{13}{10}\).

The curve is

\(x^3+2xy+8y^3=-12\).

(a) Differentiate implicitly with respect to \(x\):

\(\dfrac{\mathrm{d}}{\mathrm{d}x}(x^3)+\dfrac{\mathrm{d}}{\mathrm{d}x}(2xy)+\dfrac{\mathrm{d}}{\mathrm{d}x}(8y^3)=0\).

This gives

\(3x^2+2\left(x\dfrac{\mathrm{d}y}{\mathrm{d}x}+y\right)+24y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\).

So

\(3x^2+2x\dfrac{\mathrm{d}y}{\mathrm{d}x}+2y+24y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\).

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

\(3(-2)^2+2(-2)\dfrac{\mathrm{d}y}{\mathrm{d}x}+2(-1)+24(-1)^2\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\).

\(12-4\dfrac{\mathrm{d}y}{\mathrm{d}x}-2+24\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\).

\(10+20\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\).

Hence

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{1}{2}\).

(b) Differentiate the equation

\(3x^2+2x\dfrac{\mathrm{d}y}{\mathrm{d}x}+2y+24y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\)

again with respect to \(x\):

\(6x+2x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+2\dfrac{\mathrm{d}y}{\mathrm{d}x}+2\dfrac{\mathrm{d}y}{\mathrm{d}x}+24y^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+48y\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2=0\).

So

\(6x+2x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+4\dfrac{\mathrm{d}y}{\mathrm{d}x}+24y^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+48y\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2=0\).

Now substitute \(x=-2\), \(y=-1\), and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{1}{2}\):

\(6(-2)+2(-2)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+4\left(-\dfrac{1}{2}\right)+24(-1)^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+48(-1)\left(-\dfrac{1}{2}\right)^2=0\).

\(-12-4\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}-2+24\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}-12=0\).

\(20\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}-26=0\).

Therefore

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{26}{20}=\dfrac{13}{10}\).