Were you operating with real numbers (in a complete ordered field) the statement ##k = \frac{y}{x}## is meaningful, but you can't do that if you are strictly operating within ##\mathbb{Z}##.
The conclusion is correct that there is no such integer ##k## that produces the desired result.
Alternatively we could use the given that ##x>y##. Let's look for a candidate ##k\in\mathbb{Z}## such that ##kx=y##
Since ##k## is an integer it can be expressed as a sum ##k :=\pm ( 1+1+...+1)## (there is actually a reason why this is true, but I don't want to burst the bubble)
Looking at positive integers first
Then ##kx = y## becomes ##(1+1+...+1)x = y##. Due to the distributive property the previous can also be written as ##x + x + ... + x = y##
But we already have that ##x>y## therefore ##x+x+...+x = y## can never be true.
And if we look in the negative integers then ##-x -x ..-x = y##. Under addition, ##\mathbb{Z}## does form a group, so we can say there exists ##-(-x) = x## such that ##x+(-x) = 0## and what would follow is ##y+x+x+...+x = 0##, which can only be true if ##y>x##, violating one of our established criteria.
The initial assumption that such an integer exists is therefore incorrect. And ##x|y## only if ##0\neq |x|\leq |y|##
In actuality, such divisibility is defined in a ring with no zero terms. A ring is an algebraic structure that forms an abelian group under addition, multiplication and addition are connected via the distributive property. A ring with no zero terms (can't be too sure of the English terminology here) is a ring where no two ring elements' product is 0.
We don't know anything about "dividing both sides by.." for this problem. There is no group under multiplication.EDIT: oop, You assumed x,y to be in N, then the negative integer part is unnecessary.
Actually, this proof contains inaccuracies, as well :/
