Prove that for every real number $x$, if $x \neq 0$ then there is a unique real number $y$ such that for every number $z$, $zy = \frac{z}{x}$.
First I show that there is such that $y$, and second I check that $y$ is unique.
Existence: We let $x$ be arbitrary, and let $y = \frac{1}{x}$. We can multiply both side of the equation by $z$ to obtain $zy = \frac{z}{x}$. Thus, we found $y$ such that the equation $zy = \frac{z}{x}$ hold for all $z$ (it holds in case $z = 0$).
Uniqueness: We suppose that there are $y_1$ and $y_2$ such that $zy_1 = \frac{z}{x}$ and $zy_2 = \frac{z}{x}$. We need to show that $y_1 = y_2$. Combining the equation we obtain $zy_1 = zy_2$. Since we need to show that it holds for all $z$, then for $z \neq 0$, we get $y_1 = y_2$. But for $z = 0$, then we have possible $y_1 \neq y_2$.
So, I do not confirm the uniqueness of $y$. Am I right? Can you please check it out. (I found proof to this statement, but they not consider case when $z = 0$.)
If it has to hold for all $z$, then it has hold for both $z=0$ and $z\neq 0$. The fact that $z=0$ allows for more solution is irrelevant, because condition for $z\neq 0$ limits you to $y_1=y_2$ anyway.