Answer
$$-3$$
Work Step by Step
We are given:
$$ x^{3}+y^{3}=3 x+y-2 $$
Differentiate with respect to $x$:
\begin{align*}
3x^{2}+3y^{2}y'=3 +y'\\
y'&= \frac{3\left(1-x^{2}\right)}{3 y^{2}-1} \\
y'\bigg|_{(1,1)}&=0
\end{align*}
Now we take the second derivative:
\begin{align*}
y''&= 3\left(\frac{-2 x\left(3 y^{2}-1\right)-6 y\left(\frac{d y}{d x}\right)\left(1-x^{2}\right)}{\left(3 y^{2}-1\right)^{2}}\right)\\
&=3\left(\frac{-6 x y^{2}+2-6 y y^{\prime}\left(1-x^{2}\right)}{\left(3 y^{2}-1\right)^{2}}\right)\\
y''\bigg|_{(1,1)}&= -3
\end{align*}
