y\(^2\) - 3y > 18 = y\(^2\) - 3y - 18 > 0

y\(^2\) - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0

= (y + 3) (y - 6) > 0

Case 1 (+, +) \(\to\) (y + 3) > 0, (y - 6) > 0

=  y > -3, y > 6
By simply substituting the value of y into the inequality, y > -3 is invalid, and y < 6 is also invalid.

Therefore, y < - 3 or y > 6