Step 1: Analyze the numerator of the inequality. The given inequality is $\frac{(x+1)^2+3}{(x+3)(x-2)} < 0$. Consider the numerator: $(x+1)^2+3$. Since $(x+1)^2$ is always greater than or equal to $0$ for any real number $x$, adding $3$ to it means that $(x+1)^2+3$ is always greater than or equal to $3$. Therefore, the numerator $(x+1)^2+3$ is always positive. Step 2: Determine the sign of the denominator. For the entire fraction to be less than $0$ (negative), and since the numerator is always positive, the denominator must be negative. So, we must have $(x+3)(x-2) < 0$. Step 3: Solve the inequality for the denominator. To find when $(x+3)(x-2) < 0$, we find the roots of the quadratic expression $(x+3)(x-2) = 0$. The roots are $x = -3$ and $x = 2$. This is a parabola opening upwards. The expression $(x+3)(x-2)$ will be negative when $x$ is between its roots. Thus, the solution to $(x+3)(x-2) < 0$ is $-3 < x < 2$. Step 4: Compare with the given options. The set of values for $x$ is $\{x : -3 < x < 2\}$. This matches option D. The final answer is $\boxed{\text{D}}$.