Given:
\(\frac{x}{y} < \frac{x+3}{y-3}\)
The inequality is equivalent to:
\[ \frac{x}{y} - \frac{x+3}{y-3} < 0 \]
Combining the fractions yields:
\[ \Rightarrow \frac{x(y-3) - y(x+3)}{y(y-3)} < 0 \]
Expanding the numerator gives:
\[ \Rightarrow \frac{xy - 3x - xy - 3y}{y(y - 3)} < 0 \] \[ \Rightarrow \frac{-3(x + y)}{y(y - 3)} < 0 \]
Multiplying both sides by -1 reverses the inequality sign:
\[ \Rightarrow \frac{3(x + y)}{y(y - 3)} > 0 \]
For the inequality \( \frac{3(x + y)}{y(y - 3)} > 0 \) to hold, we analyze the signs of the numerator and denominator.
If \( y < 0 \), then \( y(y - 3) > 0 \) (product of two negatives). Therefore, for the inequality to be true, the numerator must be positive:
\[ x + y > 0 \]
This implies:
\[ x > -y \]
Considering the relationship between absolute values, we derive:
\[ |x| > |y| \]
Consequently, the final conclusion is:
\[ -x < y \]
Conclusion: Thus, option (D) is correct: "If \( y < 0 \), then \( -x < y \)."