Am I cooked for IB?

okay, this is a hard one.
start by factoring, so 3(1-(7/3)/(2/3)).
now multiply by x^2 and distribute that, but not the 3, so 3(x^2-((7x^2)/3)/((2x^2)/3)).
now, we need to set this equal to dy/dx, so dy/dx = 3(x^2-((7x^2)/3)/((2x^2)/3)).
we now have a separable differential equation, so rewrite it as dy = 3(x^2-((7x^2)/3)/((2x^2)/3))dx.
now we need to figure out the interval on which we need to evaluate this, so look at the original equation, and we get two parts, 3-7 and 2, so it is the integral from -4 to 2, so int(-4,2)(dy) = int(-4,2)(3(x^2-((7x^2)/3)/((2x^2)/3))dx). 
now we need to figure out what y is equal to, and the original equation can be reinterpreted as y = 7/2x -3.
so y](-4,2) = (x^3-21x/2)](-4,2).
y(2) = 4 and y(-4)=-17, (4-(-17) = -13-(-22), so 21=9
now /3, so 7 = 3.
take the derivate of both sides with respect to t, so 0=0, which is true. 
now multiply by x, so 0x=0x
the answer is all values of x