To solve the differential equation $y'' = 2x (y')^2$, we start by letting $p = y'$, which implies $p' = y''$. The equation becomes:

$p' = 2x p^2$

This is a separable differential equation. We can rewrite it as:

$\frac{dp}{p^2} = 2x \, dx$

Integrating both sides gives:

$-\frac{1}{p} = x^2 + C_1 \quad \Rightarrow \quad p = -\frac{1}{x^2 + C_1}$

Since $p = y'$, we now have:

$y' = -\frac{1}{x^2 + C_1}$

Integrating again, we get:

$y = -\tan^{-1}\left(\frac{x}{\sqrt{C_1}}\right) + C_2$

Thus, the general solution is:

$y = -\tan^{-1}\left(\frac{x}{C_1}\right) + C_2$

So, the correct answer is:

$y = -\tan^{-1}\left(\frac{x}{c_1}\right) + c_2$

This matches the option:

y = -tan^(-1)((x)/(c_(1)))) + c_(2).

Would you like further details on any specific part of this solution? Here are some related questions you might consider:

1. How do you derive the separation of variables for this equation?
2. What is the significance of the constants $c_1$ and $c_2$?
3. Can you explain the geometric interpretation of the solution?
4. What happens if the initial conditions are provided?
5. How does this solution compare to other types of differential equations?

Tip: Always check the initial conditions to fully determine the constants in your solution!