How exactly is implicit function theorem used in something such as this... $$\int_{x_1}^{x_2}\int_{x_1}^{x_2}\frac{\partial^2(y)}{\partial(x)^2}{dx^2}$$ It seems as if after the first round I should get something like
$$\int_{x_1}^{x_2}\left(\frac{\partial(y(x_2))}{\partial(x)}-\frac{\partial(y(x_1))}{\partial(x)}\right){dx}$$
But that doesn't make sense, you can't take a derivative of $y(x_2)$, it's a constant. If I were to use indefinite integrals, then I get $y+Cx+D$, how do I get a similar thing using definite integrals?
You don't. There should never be a double definite integral in the same variable, any definite integral "removes" the integration variable from the term completely. If you arrive at something like above, then there went something wrong before.
The analogue to a double indefinite integral occurs, when the boundary depends on a variable, something like this is allowed:
$$\int_{x_1}^{x_2} \int_{x_1}^s \frac{\partial^2 f}{\partial{x}^2}(t) dt ds = \int_{x_1}^{x_2} \left[\frac{\partial f}{\partial x}(s) - \frac{\partial f}{\partial x}(x_1) \right] ds = f(x_2)-f(x_1) - (x_2-x_1) \frac{\partial f}{\partial x}(x_1)$$
In a way, the two $x$ in your indefinite integral are actually different variables who just have been given the same name.
edit: and also something like $\frac{\partial f}{\partial x}(x_1)$ means the derivative of the function $f$ evaluated at the point $x=x_1$, you take the derivative first and then put in the point so the integral in your question is well defined, only not that useful, since you integrate over a constant.