How to tell someone that we can't imply an equation to another equation as to when multiplying $x$ both sides?

So my friends and I was trying to tell someone that this implication is wrong, he (that person) said that these equations are the same:

$$9x = 1 \Rightarrow 3x^2 = \frac x3$$

He confused why $9x = 1 \Rightarrow x={1/9, 0}$ isn't true. You see that the second one of the solutions is $x=0$, that obviously multiplying both sides with $0$ isn't wise. I was trying to explain that by fundamental theorem of algebra the first equation only has $1$ solution, but he's too obtuse to accept that and I gave up. In my real analysis class, we also had someone that doing this he thought it was legal.

If I don't mistakenly remember, it's about solving absolute equations, or whatever you call it, he suggested multiplying both sides with $0$ and that led him into wrong answers different set of solutions. My answer, however, avoiding multiplying both sides with $0$ and using basic axioms led me into correct answer which is, in my opinion, multiplying both sides with an unknown variable is really dangerous. It's really shameful I can't remember the exact problem to illustrate this.

Anyway, related to the question, how to explain this problem?

It is true that $9x=1\implies 3x^2=\frac{x}{3}$. Since $3x^2=\frac{x}{3}\iff x\in\{0,\frac{1}{9}\}$, it is also true that $9x=1\implies x\in\{0,\frac{1}{9}\}$.

However, I believe you're friend might think that this implies that $x=0$ is somehow a solution to $9x=1$. However, that is a logically flawed statement. What we would need to do is substitute $x=0$ and $x=\frac{1}{9}$ independently to determine which ones were actual solutions. And we come up with that $x=\frac{1}{9}$ is the only solution to $9x=1$.

Does ${0\times 3 = 0\times 5 = 0}$ imply ${3 = 5}$? No, since we would be dividing by $0$. Consider $$ x\times (3x) = x\times \left(\frac{1}{3}\right) $$ Just like our obviously wrong ${3=5}$ , if ${x=0}$ we cannot conclude ${3x = \frac{1}{3}}$.