Answer: (a) \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{3}{4}\) when \(x=0\).

(b) The Maclaurin series for \(y\), up to and including the term in \(x^2\), is

\(\displaystyle y=-\frac{3}{4}x+\frac{9}{64}x^2\).

We are given

\((x+1)y+(x+y+1)^3=1\), with \(y=0\) when \(x=0\).

So \(y(0)=0\).

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

\((x+1)y'+y+3(x+y+1)^2(1+y')=0.\)

Now put \(x=0\) and \(y=0\):

\(y'+3(1+y')=0.\)

So

\(4y'+3=0 \quad \Rightarrow \quad y'=-\frac{3}{4}.\)

Hence \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{3}{4}\) when \(x=0\).

(b) Differentiate the first-derivative equation again:

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

Now substitute \(x=0\), \(y=0\) and \(y'=-\frac{3}{4}\):

\(y''+2\left(-\frac{3}{4}\right)+3y''+6\left(\frac14\right)^2=0.\)

So

\(4y''-\frac32+\frac38=0,\)

\(4y''-\frac98=0,\)

\(y''=\frac{9}{32}.\)

Now use the Maclaurin expansion

\(y=y(0)+y'(0)x+\frac{y''(0)}{2}x^2+\cdots.\)

Since \(y(0)=0\), \(y'(0)=-\frac34\) and \(y''(0)=\frac{9}{32}\),

\(y=-\frac34x+\frac{1}{2}\cdot\frac{9}{32}x^2+\cdots \,=\, -\frac34x+\frac{9}{64}x^2+\cdots.\)

Therefore, up to and including the term in \(x^2\),

\(\boxed{y=-\frac34x+\frac{9}{64}x^2}.\)