I am getting $x+y=1,xy=0$, what's happening?? EDIT: Just now noticed "real numbers"

Set $x-y=p$, where $p$ is a prime. Assume on the contrary that $p \ne 3$. Then $x+y=\frac{x^2-y^2}{x-y}=\frac{q}{p}$, where $q$ is a prime, possibly the same as $p$. Thus, we get $x=\frac{p^2+q}{2p}$ and $y=\frac{q-p^2}{2p}$. Hence, $x^3-y^3=\frac{1}{8p^3}\left((p^2+q)^3-(q-p^2)^3\right)=\frac{1}{4p}\left( 3q^2+p^4 \right) \in \mathbb{N}$, and so $p \ne 3, p|q \implies p=q$ and so $x^3-y^3=\frac{p}{4}(p^2+3)=r$, where $r$ is a prime. Clearly $p=2$ does not work. Hence, $4 \nmid p$ and so $4|p^2+3$. Thus, $p|r \implies p=r$. But then we get $p=1$, absurd. Hence, $x-y=p=3$, as desired. $\blacksquare$

There exists such $x,y$ for instance set $(x,y)=\left( \frac{7}{3}, \frac{-2}{3} \right), \left( \frac{10}{3}, \frac{1}{3} \right)$. The real question is do there exist infinitely many $(x,y)$ satisfying this property?

Equivalent form of this question

Actually we only need $x-y$ and $x^3-y^3$ to be prime. Let $x=\frac{a_1}b, y=\frac {a_2}b$, and $a_1-a_2=pb$, where $p$ is a prime. Also WLOG assume $a_1, a_2$ are coprime with $b$. We have $x^3-y^3=\frac{pb(a_1^2+a_1a_2+a_2^2)}{b^3}=\frac{p(a_1^2+a_1a_2+a_2^2)}{b^2}$. But since $a_1-a_2=pb$, we have $a_1=a_2 \mod b$, so $a_1^2+a_1a_2+a_2^2=3a_1^2 \mod b$. So $b^2|3pa_1^2$, but $a_1$ and $b$ are coprime, so either $b=1$(which would make $3|x^3-y^3$, contradiction) or $b^2|3p\Rightarrow b=p=3$, QED.

@above $x,y$ aren't necessarily rational numbers. One question: Is the condition $x^2-y^2$ is a prime necessary? Can we find an example where the problem won't hold without this condition?

hansu wrote: $x,y$ aren't necessarily rational numbers. If $x^2-y^2$ is rational then $x-y$ is prime and $x+y$ is rational, so $x, y$ would be rational and the rest of my proof holds. So we don't need $x^2-y^2$ to be prime, just rational. On the other hand, $x=\frac{17}3, y=\frac8{3}$ is an example where $x-y=3, x^3-y^3=163$ are prime but $x^2-y^2=25$ is not. So, can we drop any assumption on $x^2-y^2$? We have that $x-y, x^3-y^3$ are rational, so $x^2+xy+y^2$ is rational, but $x^2+xy+y^2=(x-y)^2+3xy$, so $xy$ is rational. Now, if $x-y$ and $xy$ are rational, then the polynomial $P(\lambda)=\lambda^2-(x-y)\lambda-xy=(\lambda-x)(\lambda+y)$ has rational coefficients, and its roots are $x, -y$, so $x,y=\frac{\pm p+b\sqrt{c}}2$, where $p=x-y$ is prime, $b,c\in\mathbb{Q^+}$ and WLOG $c$ is either $1$ or square-free. (If $c=1$ then $x,y$ are rational and we're in the previous case, so assume $c\neq1$.) Then $x^3-y^3=\frac{p(p^2+3b^2c)}4$ is prime. Here we find a counterexample: put $p=2,b=1,c=2$, then $x=\frac{2+\sqrt2}2, y=\frac{-2+\sqrt2}2$, and $x-y=2, x^3-y^3=5$. Therefore, if we do not assume anything on $x^2-y^2$, then we can't conclude $x-y=3$; but the condition on $x^2-y^2$ can be weakened to just $x^2-y^2$ being rational.

Say $x-y=p.$ Evidently $x+y=\frac{q}{p}$ for some prime $q$ so we have $x=\frac{q+p^2}{2p}$ and $y=\frac{q-p^2}{2p}.$ So \[(x-y)^3=p(x^2+y^2+xy)=p(\frac{q^2}{p^2}-xy)=p(\frac{4q^2}{4p^2}-\frac{q^2-p^4}{4p^2})=\frac{3q^2+p^4}{4p}.\]We require $p\mid 3q^2+p^4,$ or $p\mid 3q^2.$ If $p\neq 3$ then $p=q,$ but $\frac{3p+p^3}{4}$ is not prime since this expression being integer necessitates $4\mid p^2+3$, but $p\frac{p^2+3}{4}$ is not prime unless $p=1,$ contradiction.