You did essentially everything right.
The parabola and line meet at $y=-2$ and $y=3$. So our area is $\int_{-2}^3 (6-y^2 -(-y))\,dy$ It is clear from your answer that you indeed ended up calculating $\int_{-2}^3 (6-y^2 +y)\,dy$
An antiderivative is given by $6y-\frac{y^3}{3}+\frac{y^2}{2}$
So the answer is $\left(6(3)-\frac{3^3}{3}+\frac{3^2}{2}\right)-\left(6(-2)-\frac{(-2)^3}{3}+\frac{(-2)^2}{2}\right)$
Your next to last term has a sign error. You have $-\frac{(-2)^3}{3}$ but when you "open" my brackets above, it should turn to $+$. (Then, when you expand $(-2)^3$, you get a $-$ again! Minus signs are evil.)
Minor Comment: You did not remove all parentheses before integrating. That left another layer of them, and may have contributed to the minus sign slip.