Consider the following system of differential equations Consider the following model. Find the general solution  

$\frac{dx}{dt}=-2y$,
$\frac{dy}{dt}=8x$

So far this is what I have: 

$$\frac{dx}{dt}=-2y$$
$$x'=-2y$$
$$y=-\frac{x'}{2}$$
$$y=-\frac{x''}{2}$$
and then
$$\frac{dy}{dt}=8x$$
$$y'=8x$$
$$-\frac{x''}{2}=8x$$
$$-x''=16x$$
$$x''+16x=0$$ 
Is this correct so far? What step should I take next?
Although not the usual way I've seen this done (usually I have seen the eigenvalue/eigenvector method done), your approach works. 

Now you need to know that the general solution to $x''+16x=0$ is $c_1 \cos(4x) + c_2 \sin(4x)$. One way to see this is to get the characteristic polynomial $\lambda^2+16$ which has roots $\pm 4i$. So from this you get a general solution of $c_1 e^{4ix} + c_2 e^{-4ix}$ for complex constants $c_1,c_2$. That can then be rearranged into the previous form.
$$\frac{dx}{dt}=-2y$$
$$\frac{dt}{dx}=-\frac{1}{2y}$$
By chain rule:
$$\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}$$
$$\frac{dy}{dx}=8x\cdot-\frac{1}{2y}$$
$$\frac{dy}{dx}=-\frac{4x}{y}$$
$$y\cdot\frac{dy}{dx}=-4x$$
$$\int y\cdot\frac{dy}{dx}dx=\int-4xdx$$
$$\int ydy=-4\int xdx$$
$$\frac{y^2}{2}+c_1=-4\cdot\frac{x^2}{2}$$
$$y^2+c_2=-4x^2$$
$$y^2+4x^2+c_2=0$$
Where $2c_1=c_2$, and $c_1$ is a constant of integration.