$K_{3,3}$ is not planar

Parameters. $K_{3,3}$: $v=6$, $e=9$, bipartite.

Euler. If planar, $f=2-6+9=5$.

Face bound. Since $K_{3,3}$ is bipartite, every cycle has length $\ge4$, so every face has $\ge4$ edges. Thus $4f\le2e$, i.e. $20\le18$, contradiction.

Conclusion. $K_{3,3}$ is not planar.