Question $1$: The equation is equivalent to $2x^2-2y^2=x-y$. The left-hand side factors as $2(x+y)(x-y)$. So our equation can be rewritten as
Thus any pair $(x,y)$ such that $x\ne y$ and $2(x+y)=1$ is a counterexample to the assertion that $x$ must be equal to $y$. This was pointed out by ncmathsadist.
Edit: It turns out that one is supposed to show that the only integer solutions have $x=y$. This follows from the above calculation, since $2(x+y)=1$ has no integer solutions.
Question $2$: We look at the question as amended. We can factor and rewrite the equation as $(x-y)(x^2+xy+y^2)=-(x-y)$. If $x\ne y$, we can cancel $x-y$, and obtain $x^2+xy+y^2=-1$.
But the equation $x^2+xy+y^2=-1$ has no real solutions, since $x^2+xy+y^2\ge 0$ for all real $x$ and $y$. One way to see this is to complete the square, getting
The right-hand side is clearly never negative for real $x$ and $y$. So it is true that the only solutions of the original equation have $x=y$.