Answer
a) $$\frac{-1}{3}$$
b) $$\frac{10}{27}$$
Work Step by Step
We are given
$$x y^{2}+y-2=0$$
a) We implicitly derive both sides:
\begin{align*}
x \cdot 2 y y^{\prime}+y^{2} \cdot 1+y^{\prime}&=0\\
y'&=-\frac{y^{2}}{2 x y+1}\\
y'\bigg|_{(1,1)}&=\frac{-1}{3}
\end{align*}
b) Now we take the second derivative:
\begin{align*}
y''&= -\frac{(2 x y+1)\left(2 y y^{\prime}\right)-y^{2}\left(2 x y^{\prime}+2 y\right)}{(2 x y+1)^{2}}\\
y''\bigg|_{(1,1)}&=-\frac{(3)\left(-\frac{2}{3}\right)-(1)\left(-\frac{2}{3}+2\right)}{3^{2}}\\
&=-\frac{-6+2-6}{27}\\
&=\frac{10}{27}
\end{align*}
