Is my (first) proof by induction correct? I am self-studying and would very much appreciate some feedback on my first ever induction proof.

Let $y_1 = 6$, and for each $n\in \mathbb{N}$ define $y_{n+1} = (2y_n - 6)/3$. Use induction to prove the sequence satisfies $y_n > -6$.
First, we'll prove the base case. Because we've been given $y_1 = 6 > -6$, our base case will be $n=2$:
$y_2 = (2y_1-6)/3 = (2(6)-6)/3 = 2 > -6.$
Next, the induction step. Assume $y_{n+1} > -6$. Let's then prove $y_{n+2} > -6$ for some $y_{n+1}$. We'll do this by contradiction, assuming $y_{n+2} = (2y_{n+1}-6)/3 < -6$. Observe:
$$(2y_{n+1}-6)/3 < -6,$$
$$2y_{n+1}-6 < -18,$$
$$2y_{n+1} < -12,$$
$$y_{n+1} < -6,$$
which contradicts our initial assumption that $y_{n+1} > -6$. Thus, $y_{n+2} > -6$.

You can still use the base case $n=1$ despite it holding automatically. In general for induction proofs it is more natural to write 'assume for $n$, then true for $n+1$', although it is essentially the same as what you have done. Having said that, for the proof by contradiction to work you need to start by assuming the precise opposite of what you are trying to prove, so assume $y_{n+2}\leq -6$ instead and you're good to go.
A good way to think of induction proofs is like a 'domino effect'. Check that it works for the first domino ($n=1$), and that if it works for $n$ then it must work for $n+1$. It then must work for all $n\geq 1$ by induction.
A cleaner proof: Clearly $y_1>-6$. If $y_n>-6$, then $y_{n+1}=\frac{1}{3}(2y_n-6)>-6$.
$y_1>-6$ and ($y_n>-6\implies y_{n+1}>-6$) gives that $y_n>-6$ by induction for all $n\geq 1$.
