Answer: (a) \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac12\) at \(\left(4,\frac13\right)\).

(b) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{29}{4}\) at \(\left(4,\frac13\right)\).

The curve is

\(9y^2-3\sinh^{-1}(xy)=1-3\ln 3\).

Differentiating implicitly with respect to \(x\),

\(18yy'-3\dfrac{xy'+y}{\sqrt{1+x^2y^2}}=0\).

(a) At \(\left(4,\frac13\right)\),

\(18\left(\frac13\right)y'-3\dfrac{4y'+\frac13}{\sqrt{1+16\left(\frac13\right)^2}}=0\)

\(6y'-3\dfrac{4y'+\frac13}{5/3}=0\)

\(6y'-\frac95\left(4y'+\frac13\right)=0\).

So

\(30y'-36y'-3=0\)

\(-6y'-3=0\)

and hence

\(y'=-\frac12\).

(b) Differentiate

\(18yy'-3\dfrac{xy'+y}{\sqrt{1+x^2y^2}}=0\)

again.

For the first term,

\(\dfrac{\mathrm{d}}{\mathrm{d}x}(18yy')=18yy''+18(y')^2\).

For the second term, using the product rule on \((xy'+y)(1+x^2y^2)^{-1/2}\),

\(\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{xy'+y}{\sqrt{1+x^2y^2}}\right)=\frac{xy''+2y'}{\sqrt{1+x^2y^2}}-\frac{(xy'+y)^2(xy)}{(1+x^2y^2)^{3/2}}.\)

So

\(18yy''+18(y')^2-3\left(\dfrac{xy''+2y'}{\sqrt{1+x^2y^2}}-\dfrac{(xy'+y)^2(xy)}{(1+x^2y^2)^{3/2}}\right)=0\).

Now substitute \(x=4\), \(y=\frac13\), \(y'=-\frac12\):

\(1+x^2y^2=\frac{25}{9},\quad \sqrt{1+x^2y^2}=\frac53,\quad xy'+y=4\left(-\frac12\right)+\frac13=-\frac53,\quad xy=\frac43.\)

This gives

\(18\left(\frac13\right)y''+18\left(\frac14\right)-3\left(\frac{4y''-1}{5/3}-\frac{\left(\frac{25}{9}\right)\left(\frac43\right)}{(25/9)(5/3)}\right)=0\).

Equivalently,

\(6y''+\frac92-3\left(\frac35(4y''-1)-\frac45\right)=0\).

Simplifying,

\(6y''+\frac92-\frac{36}{5}y''+\frac{21}{5}=0\)

\(-\frac65y''+\frac{87}{10}=0\).

Hence

\(y''=\frac{29}{4}\).

Therefore, at \(\left(4,\frac13\right)\),

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{29}{4}\).