Answer: (a) \(\displaystyle \frac{dy}{dx}=-\frac{2}{3}\) at \((0,-2)\).

(b) \(\displaystyle \frac{d^2y}{dx^2}=-\frac{4}{27}\) at \((0,-2)\).

The curve is

\((x+1)y+y^2=2\).

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

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

Using the product rule,

\((x+1)\frac{dy}{dx}+y+2y\frac{dy}{dx}=0\).

So

\((x+1+2y)\frac{dy}{dx}+y=0\),

hence

\(\displaystyle \frac{dy}{dx}=-\frac{y}{x+1+2y}\).

At \((0,-2)\),

\(\displaystyle \frac{dy}{dx}=-\frac{-2}{0+1+2(-2)}=\frac{2}{-3}=-\frac{2}{3}\).

Therefore \(\displaystyle \frac{dy}{dx}=-\frac{2}{3}\).

(b) Differentiate

\((x+1)\frac{dy}{dx}+y+2y\frac{dy}{dx}=0\)

again:

\(\frac{d}{dx}\left((x+1)\frac{dy}{dx}\right)+\frac{d}{dx}(y)+\frac{d}{dx}\left(2y\frac{dy}{dx}\right)=0\).

This gives

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

So

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

Now substitute \(x=0\), \(y=-2\), and \(\frac{dy}{dx}=-\frac{2}{3}\):

\((1-4)\frac{d^2y}{dx^2}+2\left(-\frac{2}{3}\right)+2\left(-\frac{2}{3}\right)^2=0\).

\(-3\frac{d^2y}{dx^2}-\frac{4}{3}+\frac{8}{9}=0\).

\(-3\frac{d^2y}{dx^2}-\frac{4}{9}=0\).

\(-3\frac{d^2y}{dx^2}=\frac{4}{9}\).

\(\displaystyle \frac{d^2y}{dx^2}=-\frac{4}{27}\).