Solve the inequality $f(x)>0$, where $f(x)=x^2(x+3)$, by using the graph of the function.
The solution set for $f(x)>0$ is (Type your answer in interval notation.)

Step 1 ：Given the inequality $f(x)>0$, where $f(x)=x^2(x+3)$, we are looking for the values of $x$ where the function $f(x)$ is positive.

Step 2 ：The function $f(x)=x^2(x+3)$ is a cubic function and its graph will cross the x-axis at the roots of the equation $f(x)=0$. The roots of the equation are $x=0$ and $x=-3$.

Step 3 ：We can divide the number line into three intervals based on these roots: $(-\mathrm{\infty },-3)$, $(-3,0)$, and $(0,\mathrm{\infty })$. We can then test a number from each interval to see if the function is positive or negative in that interval.

Step 4 ：For the interval $(-\mathrm{\infty },-3)$, we can test $x=-4$. For the interval $(-3,0)$, we can test $x=-1$. For the interval $(0,\mathrm{\infty })$, we can test $x=1$.

Step 5 ：The function $f(x)$ is negative for $x$ in the interval $(-\mathrm{\infty },-3)$ and positive for $x$ in the intervals $(-3,0)$ and $(0,\mathrm{\infty })$.

Step 6 ：Therefore, the solution set for the inequality $f(x)>0$ is $(-3,0)\cup (0,\mathrm{\infty })$.

Step 7 ：$\boxed{{\displaystyle \text{The solution set for the inequality }f(x)>0\text{ is }(-3,0)\cup (0,\mathrm{\infty })}}$