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

Step 1 ：The function $f(x)=x^2+6x+9$ 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 $f(x)=a{x}^2+bx+c$ is $-\frac{b}{2a}$. In this case, $a=1$ and $b=6$, so the x-coordinate of the vertex is $-\frac{6}{2\ast 1}=-3$.

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

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