Let $\sigma(x)$ be the sum of divisors of $x$, and denote the abundancy index of $x$ by $I(x) = \sigma(x)/x$.
Consider the number $y^2 \in \mathbb{N}$, and suppose that I know that $I(y^2) < 4/3$.
Is the following a valid proof?
Proposition
If $I(y^2) < 4/3$, then $3 \nmid y$.
Proof of Proposition
Assume that $I(y^2) < 4/3$ is true. Suppose to the contrary that $3 \mid y$. Then $y^2 = {3^{2\alpha}}{x^2}$ for some $\alpha \geq 1$, and note that $\gcd(3, x) = 1$. Therefore, we obtain $$\dfrac{4}{3} > I(y^2) = I\left(3^{2\alpha}\right)I(x^2) \geq I(3^2) = \dfrac{1+3+9}{9}=\dfrac{13}{9},$$ where we have used the facts that $I(m) \geq I(n)$ if $n \mid m$, and that $I(k) \geq 1 \hspace{0.05in}\forall k \in \mathbb{N}$. This is a contradiction.
QED
For a concrete example, let $y = 5$. Then we obtain $$\dfrac{4}{3} > I(y^2) = I(5^2) = \dfrac{1+5+25}{25} = \dfrac{31}{25},$$ which is true. Note that $3 \nmid y = 5$.
For an alternative proof, take into consideration that if the proposition is true, then the contrapositive is true.
Contrapositive
If $3 \mid y$, then $I(y^2) \geq 4/3$.
Proof of Contrapositive
Assume that $3 \mid y$. Then $y$ can be written in the form $y = {3^{\alpha}}x$, where $\alpha \geq 1$ and $\gcd(3,x)=1$. Consequently, $$I(y^2) = I\left({3^{2\alpha}}{x^2}\right) = I(3^{2\alpha})I(x^2) \geq I(3^2) = \dfrac{13}{9} > \dfrac{4}{3}.$$
QED
I think your proof is fine.
Proceeding by contrapositive would also work well. If $3\mid y$ then $9\mid y^2$ and $I(y^2)\ge I(9) = \frac{13}{9}>\frac43$.