$\displaystyle y^2+y=\frac{x^3}{3}+c$ Im not sure how to get y on its own at this stage
you have a quadratic equation in y. it's hard to factor of course, but you can always complete the square, or use the quadratic formula (here, your constant term is $\displaystyle -(x^3/3 + C)$)
c=1? do i need to solve for y to answer this question?
actually, no, C is not 1
you do not need to solve for y to find C, however, when a problem says "solve the differential equation", they usually expect you to end up with y = ... or if x is your output variable, x = ...
y(0) = -1 means when x = 0, y = -1. with this you can find C
the quadratic formula says the solution to a polynomial equation of the form $\displaystyle ay^2 + by + c = 0$ is $\displaystyle y = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$
when I got c=1, I got it because I let -1=Left side $\displaystyle (y^2$+y). So I guess I should have done$\displaystyle (-1)^2$-(-1)=$\displaystyle \frac{(0)}{3}^3$+C. But this would give C to be 0