Answer: (a) \(\displaystyle \frac{dy}{dx}=-\frac16\) at \(\left(2,\frac12\right)\).

(b) \(\displaystyle \frac{d^2y}{dx^2}=\frac{11}{108}\) at \(\left(2,\frac12\right)\).

The curve is

\(4y^2+4\ln(xy)=1\).

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

\(\frac{d}{dx}(4y^2)+\frac{d}{dx}(4\ln(xy))=0\).

Now

\(\frac{d}{dx}(4y^2)=8yy'\),

and, using the chain rule with \(u=xy\),

\(\frac{d}{dx}(4\ln(xy))=4\cdot\frac{1}{xy}\cdot\frac{d}{dx}(xy)=4\cdot\frac{xy'+y}{xy}=4y'y^{-1}+4x^{-1}.\)

So

\(8yy'+4y^{-1}y'+4x^{-1}=0.\)

Substitute \(x=2\) and \(y=\frac12\):

\(8\left(\frac12\right)y'+4\left(\frac12\right)^{-1}y'+4(2)^{-1}=0.\)

Hence

\(4y'+8y'+2=0\),

\(12y'=-2\),

so

\(\displaystyle y'=-\frac16.\)

(b) Differentiate

\(8yy'+4y^{-1}y'+4x^{-1}=0\)

again:

\(\frac{d}{dx}(8yy')+\frac{d}{dx}(4y^{-1}y')+\frac{d}{dx}(4x^{-1})=0.\)

Using the product rule,

\(\frac{d}{dx}(8yy')=8(y')^2+8yy''\),

\(\frac{d}{dx}(4y^{-1}y')=4\big((-y^{-2}y')y'+y^{-1}y''\big)=-4y^{-2}(y')^2+4y^{-1}y''\),

and

\(\frac{d}{dx}(4x^{-1})=-4x^{-2}.\)

So

\(8(y')^2+8yy''-4y^{-2}(y')^2+4y^{-1}y''-4x^{-2}=0.\)

Now substitute \(x=2\), \(y=\frac12\), and \(y'=-\frac16\):

\(8\left(-\frac16\right)^2+8\left(\frac12\right)y''-4\left(\frac12\right)^{-2}\left(-\frac16\right)^2+4\left(\frac12\right)^{-1}y''-4(2)^{-2}=0.\)

This gives

\(\frac{2}{9}+4y''-\frac{4}{9}+8y''-1=0,\)

so

\(12y''-\frac{11}{9}=0.\)

Hence

\(\displaystyle y''=\frac{11}{108}.\)