Where is the function $f(x)=x^{2}+6 x+9$ increasing? Where is it decreasing?

Step 1 ：The function \(f(x)=x^{2}+6 x+9\) is a quadratic function. The graph of a quadratic function is a parabola.

Step 2 ：If the coefficient of \(x^{2}\) is positive, the parabola opens upwards, and if it is negative, the parabola opens downwards. In this case, the coefficient of \(x^{2}\) is positive, so the parabola opens upwards.

Step 3 ：This means that the function is decreasing to the left of the vertex and increasing to the right of the vertex.

Step 4 ：The x-coordinate of the vertex of a parabola given in the form \(f(x)=ax^{2}+bx+c\) is \(-\frac{b}{2a}\). In this case, \(a=1\) and \(b=6\), so the x-coordinate of the vertex is \(-\frac{6}{2*1}=-3\).

Step 5 ：Therefore, the function is decreasing for \(x<-3\) and increasing for \(x>-3\).

Step 6 ：Final Answer: The function \(f(x)=x^{2}+6 x+9\) is increasing for \(x>-3\) and decreasing for \(x<-3\). Therefore, the final answer is \(\boxed{x>-3, x<-3}\).