smc

We have two cases to consider: $x$ is positive or $x$ is negative. If $x$ is positive, we have $x+y=3$ and $xy+x^3=0$ Solving for $y$ in the top equation gives us $3-x$. Plugging this in gives us: $x^3-x^2+3x=0$. Since we're told $x$ is not zero, we can divide by $x$, giving us: $x^2-x+3=0$ The discriminant of this is $(-1{)}^2-4(1)(3)=-11$, which means the equation has no real solutions. We conclude that $x$ is negative. In this case $-x+y=3$ and $-xy+x^3=0$. Negating the top equation gives us $x-y=-3$. We seek $x-y$, so the answer is $\boxed{-3}$