jmc

First we'll simplify that complicated expression. We attempt to factor the numerator of the left side: $$\begin{array}{rl}p{q}^2+p^{2}q+3q^2+3pq& =q(pq+p^2+3q+3p)\\ & =q[p(q+p)+3(q+p)]\\ & =q(p+3)(q+p).\end{array}$$Substituting this in for the numerator in our inequality gives $$\frac{3q(p+3)(p+q)}{p+q}>2p^{2}q.$$We note that left hand side has $p+q$ in both the numerator and denominator. We can only cancel these terms if $p+q\ne 0.$ Since we're looking for values of $p$ such that the inequality is true for all $q>0,$ we need $p\ge 0$ so that $p+q\ne 0.$

Also because this must be true for every $q>0$, we can cancel the $q$'s on both sides. This gives $$\begin{array}{rl}3(p+3)& >2p^2\Rightarrow \\ 3p+9& >2p^2\Rightarrow \\ 0& >2p^2-3p-9.\end{array}$$Now we must solve this quadratic inequality. We can factor the quadratic as $2p^2-3p-9=(2p+3)(p-3)$. The roots are $p=3$ and $p=-1.5$. Since a graph of this parabola would open upwards, we know that the value of $2p^2-3p-9$ is negative between the roots, so the solution to our inequality is $-1.5<p<3.$ But we still need $0\le p,$ so in interval notation the answer is $\boxed{[0,3)}$.