Answer: (a) At the point \((-1,1)\), \(\frac{\mathrm{d}y}{\mathrm{d}x}=11\).

(b) At the point \((-1,1)\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=900\).

The curve is

\(xy^3-4x^3y=3\).

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

\(\frac{\mathrm{d}}{\mathrm{d}x}(xy^3)-\frac{\mathrm{d}}{\mathrm{d}x}(4x^3y)=0\).

Using the product rule on both terms,

\(3xy^2y'+y^3-4(x^3y'+3x^2y)=0\).

So

\((3xy^2-4x^3)y'=12x^2y-y^3\).

Now substitute \((x,y)=(-1,1)\):

\((3(-1)(1)^2-4(-1)^3)y'=12(-1)^2(1)-1^3\).

\((-3+4)y'=12-1\), so \(y'=11\).

Hence \(\frac{\mathrm{d}y}{\mathrm{d}x}=11\) at \((-1,1)\).

(b) Differentiate

\((3xy^2-4x^3)y'=12x^2y-y^3\).

Using the product rule,

\((3xy^2-4x^3)y''+y'(6xyy'+3y^2-12x^2)=24xy+12x^2y'-3y^2y'\).

Now substitute \(x=-1\), \(y=1\), \(y'=11\):

\((3(-1)(1)^2-4(-1)^3)y''+11\bigl(6(-1)(1)(11)+3(1)^2-12(-1)^2\bigr)=24(-1)(1)+12(-1)^2(11)-3(1)^2(11).\)

This gives

\(y''+11(-75)=75\).

So \(y''-825=75\), hence \(y''=900\).

Therefore \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=900\) at \((-1,1)\).