A simple proof by induction $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N}$ Verify by induction that $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N} \quad \forall n \ge 1 \in \mathbb{N}$
Basis: $P(1) \Rightarrow \frac{33}{11} \in \mathbb{N}$.
Induction: if the statement holds for some $n$ $\Rightarrow$ holds for $n+1$
I would appreciate some small hints because I'm stuck at $P(n+1) = \frac{6^{2(n+1)} - 3^{n+1}}{11}$

We have $P(n)=\frac{6^{2n}-3^n}{11} \in \mathbb{N}$.
You have proven that the base case is true.
Now, assume true for $n=k$.
$$6^{2k}-3^k=11p \tag{1}$$
Where $p \in \mathbb{N}$.
True for $n=k+1$.
$6^{2(k+1)}-3^{k+1}=6^{2k+2}-3^{k+1}=36 \cdot 6^{2k}-3 \cdot 3^k \tag{2}$
Now, here comes the trick.
We can substitute from $6^{2k}$ from equation $(1)$ into equation $(2)$.
$$36 \cdot (11p+3^k)-3 \cdot 3^k=11\cdot 36p+36\cdot 3^k-3 \cdot 3^k=11 \cdot 36p-33 \cdot 3^k \tag{3}$$
Since $11\cdot 36p$ is obviously divisible by $11$, and $33 \cdot 3^k$ can be divided by $11$ to give $3 \cdot 3^k$, the whole of equation $(3)$ is divisible by $11$.
Hence, we are done.
