Continuity function with two variables

Question

Let $E=\{(x,y) \in R^2 : x > y \} \cup \{(0,0)\}$. I have to show that there is no $\alpha \in R$ such that $f : E \rightarrow R$ defined by : $$f(x,y)=\frac{e^{-\frac{1}{x^2+y^2}}}{\sqrt{x-y}} \quad \text{if }x>y $$ $$f(x,y)=\alpha \quad \text{if }x=y=0$$ is continuous on $(0,0)$. My attempt : suppose that there exist such $\alpha$ and that $f$ is continuous on $(0,0)$. Then $$\forall \epsilon > 0, \quad \exists \delta >0 \quad |\quad \sqrt{x^2+y^2}<\delta \Rightarrow \mid f(x,y)-\alpha | < \epsilon $$ We have $\frac{1}{x^2+y^2}>\frac{1}{\delta^2}>0$ so $$\mid e^{-\frac{1}{\sqrt{x^2+y^2}}}(x-y)^{-1/2}-\alpha \mid < \mid e^{-\frac{1}{\delta^2}}(x-y)^{-1/2}\mid + \mid \alpha \mid$$ I would like to get something absurd but i’m stuck... Thank you for your help !