Question - Identifying the Graphical Representation of an Inequality Solution

First, find the critical points by solving \( 3x + 1 > 3 \) and \( 3x + 1 \leq 3 \):

\[ 3x > 2 \]

\[ x > \frac{2}{3} \]

And

\[ 3x \leq 2 \]

\[ x \leq \frac{2}{3} \]

From the inequality \( 3x + 1 > 3(3x + 1)^2 \), we note that the solution set represents values for x where the linear expression \( 3x + 1 \) is greater than its square \( (3x + 1)^2 \) implying it lies above the parabola \( y = x^2 \) in the coordinate plane for certain intervals. This typically occurs when \( 3x + 1 \) is between 0 and 1 where a number is greater than its square.

Solving for this we get:

\[ 0 < 3x + 1 < 1 \]

\[ -\frac{1}{3} < x < 0 \]

We have two intervals where the inequality might hold true: \( x < -\frac{1}{3} \) or \( -\frac{1}{3} < x < 0 \). But from plotting or analyzing the inequality, we can infer that the interval that satisfies the original inequality is \( -\frac{1}{3} < x < \frac{2}{3} \), which is option (C).