Answer: (i) \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3}{4}\).

(ii) \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{1}{16}\).

Differentiate the curve \(x^3-2y^3=3xy\) implicitly with respect to \(x\).

For the left-hand side, \(\frac{\mathrm{d}}{\mathrm{d}x}(x^3)=3x^2\) and \(\frac{\mathrm{d}}{\mathrm{d}x}(-2y^3)=-6y^2\frac{\mathrm{d}y}{\mathrm{d}x}\).

For the right-hand side, use the product rule:

\(\frac{\mathrm{d}}{\mathrm{d}x}(3xy)=3y+3x\frac{\mathrm{d}y}{\mathrm{d}x}\).

So

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

At \(P(2,1)\),

\(12-6\frac{\mathrm{d}y}{\mathrm{d}x}=3+6\frac{\mathrm{d}y}{\mathrm{d}x}\),

hence \(9=12\frac{\mathrm{d}y}{\mathrm{d}x}\), so

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3}{4}.\)

Now differentiate the equation \(3x^2-6y^2y'=3y+3xy'\) again, where \(y' = \frac{\mathrm{d}y}{\mathrm{d}x}\) and \(y'' = \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\).

Differentiate each term:

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

This gives

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

Simplify:

\(6x-12y(y')^2-6y^2y''=6y'+3xy''.\)

Substitute \(x=2\), \(y=1\), and \(y'=\frac34\):

\(12-12\left(\frac34\right)^2-6y''=6\left(\frac34\right)+6y''.\)

Since \(12\left(\frac34\right)^2=12\cdot\frac{9}{16}=\frac{27}{4}\), this becomes

\(12-\frac{27}{4}-6y''=\frac{9}{2}+6y''.\)

So

\(\frac{21}{4}-6y''=\frac{18}{4}+6y''\),

hence \(\frac{3}{4}=12y''\).

Therefore

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{1}{16}.\)