smc

From the initial equation $\frac{x}{y}=x-y$, we can solve for $y$ in terms of $x$ as follows: $$\begin{array}{rl}\frac{x}{y}& =x-y\\ x& =xy-y^2\\ y^2-xy+x& =0\\ y=\frac{x\pm \sqrt{x^2-4x}}{2}& \end{array}$$ Because $y$ is real, the discriminant of the above expression for $y$ must be $\ge 0$, so $x^2-4x=x(x-4)\ge 0$. This is true when $x\ge 4$ and $x\le 0$, so we choose answer $\boxed{x\ge 4\text{ or }x\le 0}$.