Answer: (a) \(\displaystyle \frac{dy}{dx}=\frac{1}{17}\) at \((-4,3)\).

(b) \(\displaystyle \frac{d^2y}{dx^2}=-\frac{96}{289}\) at \((-4,3)\).

Given

\(4y^3+(x+y)^6=109\).

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

\(12y^2\frac{dy}{dx}+6(x+y)^5\left(1+\frac{dy}{dx}\right)=0\).

At \((-4,3)\), we have \(y=3\) and \(x+y=-1\). So

\(12(3^2)\frac{dy}{dx}+6(-1)^5\left(1+\frac{dy}{dx}\right)=0\).

Hence

\(108\frac{dy}{dx}-6\left(1+\frac{dy}{dx}\right)=0\)

\(108\frac{dy}{dx}-6-6\frac{dy}{dx}=0\)

\(102\frac{dy}{dx}=6\)

\(\displaystyle \frac{dy}{dx}=\frac{1}{17}\).

(b) Divide the first derivative equation by \(6\):

\(2y^2y'+(x+y)^5(1+y')=0\).

Differentiate again:

\(\frac{d}{dx}(2y^2y')+\frac{d}{dx}\big((x+y)^5(1+y')\big)=0\).

Using the product rule,

\(2\big(2y(y')^2+y^2y''\big)+5(x+y)^4(1+y')^2+(x+y)^5y''=0\).

So

\(4y(y')^2+2y^2y''+5(x+y)^4(1+y')^2+(x+y)^5y''=0\).

Now substitute \(y=3\), \(x+y=-1\), and \(y'=\frac{1}{17}\):

\(4(3)\left(\frac{1}{17}\right)^2+2(3^2)y''+5(-1)^4\left(1+\frac{1}{17}\right)^2+(-1)^5y''=0\).

This gives

\(\frac{12}{289}+18y''+5\left(\frac{18}{17}\right)^2-y''=0\)

\(\frac{12}{289}+17y''+\frac{1620}{289}=0\)

\(17y''+\frac{1632}{289}=0\).

Since \(\frac{1632}{289}=\frac{96}{17}\),

\(17y''+\frac{96}{17}=0\).

Therefore

\(\displaystyle y''=-\frac{96}{289}\).