Answer: At the point \((0,2)\), \(\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \dfrac{9}{8}\).

Differentiate the equation \(x^{3}-3xy+y^{2}=4\) implicitly with respect to \(x\).

Using the product rule on \(xy\) and the chain rule on \(y^2\),

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

Now substitute the point \((0,2)\):

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

\(-6+4\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\).

Hence \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{3}{2}\).

Differentiate the first derivative equation again:

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

Collecting terms gives

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

Now substitute \(x=0\), \(y=2\), and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{3}{2}\):

\(0-6\cdot\dfrac{3}{2}+2\left(\dfrac{3}{2}\right)^2+4\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=0\).

So

\(-9+\dfrac{9}{2}+4\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=0\).

Thus

\(4\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{9}{2}\),

and therefore

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{9}{8}\).