dy dt2 dt d'y 8x = 0 - dt?

Given that,

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}+\frac{dy}{dt}=0 \\ \frac{d^{2}y}{d{t}^2}-8x=0 \end{gathered}$$

Now take the derivative for the first equation with respect to $t$,

$$\begin{gathered} \frac{d^3}{d{t}^3}x+\frac{d^{2}y}{d{t}^2}=0 \\ x\text{'}\text{'}\text{'}+8x=0 \end{gathered}$$

Since $\frac{d^{2}y}{d{t}^2}=8x$.

So for $x\text{'}\text{'}\text{'}+8x=0$ take $m=\frac{d}{dt}$.

Then we get,

$$\begin{gathered} m^3+8=0 \\ \left(m+2\right)\left(m^2-2m+4\right)=0 \end{gathered}$$

Where $m=-2$ is one of the solution for the cubic equation now to find the roots for the quadratic equation,

$m^2-2m+4=0$ 

If the quadratic is of the form $a{x}^2+bx+c=0$ then the roots are, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

So, the roots for $m^2-2m+4=0$ is 

$\begin{array}{rcl}m& =& \frac{2\pm \sqrt{4-16}}{2}\\ & =& 1\pm \frac{i2\sqrt{3}}{2}\\ & =& 1\pm i\sqrt{3}\end{array}$