To factorize the expression \(x^{2} + 2xy + y^{2} + 3x + 3y - 18\) completely, we can first group the terms as follows:

\((x^{2} + 2xy + y^{2}) + (3x + 3y) - 18\)

The first group \((x^{2} + 2xy + y^{2})\) is a perfect square trinomial and can be factored as \((x+y)^{2}\).

The second group \((3x + 3y)\) can be factored by taking out the common factor of 3, giving us \(3(x+y)\).

So, the expression becomes:

\((x+y)^{2} + 3(x+y) - 18\)

Now, we can use the technique of completing the square to factorize this expression further. We can rewrite the expression as follows:

\((x+y)^{2} + 3(x+y) + \frac{9}{4} - 18 - \frac{9}{4}\)

\((x+y+\frac{3}{2})^{2} - \frac{81}{4}\)

Now, we can take the square root of both sides:

\(x+y+\frac{3}{2} = \pm \frac{\sqrt{81}}{2}\)

\(x+y+\frac{3}{2} = \pm \frac{9}{2}\)

Solving for \(x+y\), we get two solutions:

\(x+y = -6\) or \(x+y = 3\)

So, the expression \(x^{2} + 2xy + y^{2} + 3x + 3y - 18\) can be factored completely as \((x+y+6)(x+y-3)\).