Existence of normal subgroups for a group of order $36$

Assume that $G$ has $4$ Sylow $3$-groups, as you noted $n_3=(1+3k)|4, n_3\neq1$. Defining the conjugation action on the set of these $4$ Sylow $3$-groups, we have the induced homomorphism $\phi: G\longrightarrow S_4$. So $\frac{G}{\ker\phi}\hookrightarrow S_4$, but $3^2$ divides |G|=36 and does not divide $|S_4| = 24$, so $\ker\phi\neq 1$ and of course $\ker\phi\neq G$. This means that there must be non-trivial kernel, which is a non-trivial normal subgroup of G.

The normal subgroup of order 3 in a group of order 36 is guaranteed. By Sylow theorem the subgroup of order 9 is possible. Also index of the subgroup is 4. And 36 does not divide 4!=24. Hence there is a non trivial normal subgroup contained in a subgroup of order 9.By lagrange theorem those normal subgroup order is 3. Hence the normal subgroup of order 3 exist in group having order 36.