You have
$\displaystyle \begin{cases}x + y + z = 3 \\ x^2 + y^2 + z^2 = 9 \\ xyz = -2\end{cases}$
From equation 1:
$\displaystyle x = 3 - y - z$, so
$\displaystyle \begin{cases}(3 - y - z)^2 + y^2 + z^2 = 9\\ (3 - y - z)yz = -2\end{cases}$
$\displaystyle \begin{cases}(3 - y - z)^2 + y^2 + z^2 = 9 \\ zy^2 + (z^2 - 3z)y - 2 = 0\end{cases}$
Since the final equation is now a quadratic in $\displaystyle y$, you can solve for $\displaystyle y$ in terms of $\displaystyle z$ using the quadratic formula.
Everything else should piece together after that.