Prove there is a constant $c > 0$ s.t. for every sufficiently large n, there exists a graph with n vertices and at least $cn^{3/2}$ edges, but no K3,3-subgraph.
Hint: let p be a suitably chosen function of n, such that G(n, p) in expectation has at least on the order of $n^{3/2}$ edges. Then compute the expected number of K3,3-subgraphs. Lastly, delete one edge from each K3,3-subgraph.
I have the intuition to set $p(n)=1/n^{2/3}$ and I did compute the expected number of K3,3-subgraphs of a random graph with n vertices: there are $\frac 1 2 {n \choose 3} {n-3 \choose 3}$ unordered pair of disjoint subsets of 3 vertices, you simply want the 9 edges to be present. Thus, the expected number is $\frac 1 2 {n \choose 3} {n-3 \choose 3}p^9$, which shows $p(n)=1/n^{2/3}$ should be correct.
However, I dont know how to use the result/intuition I got above and the hint to get a full proof. Any help is appreciated.
Your expression shows that there are the order of $n^6\ K_{3,3}$ subgraphs. If you choose $p=n^{-1/2}$ the number of these that are found in the random graph is of order $n^6 \cdot (n^{-1/2})^6 \sim n^{3/2}$. You also have the order of $n^2\cdot n^{-1/2}=n^{3/2}$ expected edges. If the constant on the number of expected edges is higher than the constant on the number of $K_{3,3}$ subgraphs, you can delete one edge per subgraph and still have of order $n^{3/2}$ edges left. $c$ can be the difference of these constants or smaller.