The first thing I would do is graph the equations. The lower boundary, y= -3x, is a straight line through the origin with slope -3, the upper boundary, y= 4- x^2, is a parabola with vertex at (0, 4) opening downward. The second thing I would do is determine the coordinates of the points where those boundaries intersect. y= -3x= 4- x^2 gives x^2- 3x- 4= 0. That factors easily as (x- 4)(x+ 1)= 0 so they intersect at (-1, 3) and (4, -12).
Further I see that if I were to draw horizontal lines across that region, corresponding to different values of y, the left and right end points, the limits on the x integrals, change. For y less than or equal to 3, the left end point is on the line y= -3x. For y greater than or equal to 3, the left endpoint is on the left side of the parabola y= 4- x^2. That tells me that I have to break the integral into two parts, one for y from -12 to 3 and the other for y from 3 to 4.
The line y= -3x "inverts" to x= -y/3. The parabola y= 4- x^2 doesn't have a true "inverse". x^2= 4- y so x= (4- y)^(1/2), the right branch, or x= -(4- y)^(1/2), the left branch.
Reversing the order of integration we need to break this into two separate double integrals:
\(\displaystyle \int_{y= -12}^3 \int_{x= -y/3}^{\sqrt{4- y}} f(x,y) dxdy+ \int_{y= 3}^4\int_{x= -\sqrt{4- y}}^{\sqrt{4- y}} f(x,y) dxdy\).