Answer: At the point \(A(-1,1)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-4\).

The value of \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) at \(A\) is \(-42\).

Differentiate the equation \(2xy^2+3x^2y=1\) with respect to \(x\), treating \(y\) as a function of \(x\).

For \(2xy^2\), use the product rule:

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

For \(3x^2y\), use the product rule:

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

So

\(2y^2+4xy\dfrac{\mathrm{d}y}{\mathrm{d}x}+6xy+3x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\).

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

\(2(1)^2+4(-1)(1)\dfrac{\mathrm{d}y}{\mathrm{d}x}+6(-1)(1)+3(-1)^2\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\),

which gives

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

Hence \(-4-\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\), so

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

Differentiate the first derivative equation again:

\(2y^2+4xyy'+6xy+3x^2y'=0\), where \(y' = \dfrac{\mathrm{d}y}{\mathrm{d}x}\).

Term by term:

• \(\dfrac{\mathrm{d}}{\mathrm{d}x}(2y^2)=4yy'\)
• \(\dfrac{\mathrm{d}}{\mathrm{d}x}(4xyy')=(4y+4xy')y' + 4xyy''\)
• \(\dfrac{\mathrm{d}}{\mathrm{d}x}(6xy)=6y+6xy'\)
• \(\dfrac{\mathrm{d}}{\mathrm{d}x}(3x^2y')=6xy'+3x^2y''\)

So

\(4yy' + (4y+4xy')y' + 4xyy'' + 6y + 6xy' + 6xy' + 3x^2y'' = 0\).

Substitute \(x=-1\), \(y=1\), and \(y'=-4\):

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

Hence

\(-16 - 80 - 4y'' + 6 + 24 + 24 + 3y'' = 0\).

So \(-42 - y'' = 0\), giving

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -42\).