Using epsilon-delta to prove a multivariable limit

Question

I'm trying to find $\lim_{x,y \to 0,0} f(x,y)$ where $f(x,y)=\frac{e^\frac{-1}{\sqrt{x^2+y^2}}}{e^{x^2+y^2}-1}$ I've tried converting to polar coordinates but it doesn't seem to be a helpful simplification. Graphically, it seems that the limit is 0 from all possible paths, and I tried to construct an epsilon-delta proof to prove this. I'm just trying to figure out if I'm on the right track.
Let $\epsilon>0$ be given, find a $\delta>0$ such that if $|{(x,y)-(0,0)}|<\delta$ implies $|{f(x,y)-L}|<\epsilon$, Assuming $L=0$. We know that $\sqrt{x^2+y^2}<\delta\rightarrow x^2+y^2<\delta^2$ which implies $\frac{1}{e^{x^2+y^2}-1}<\frac{1}{e^{\delta^2}-1}$ $|{f(x,y)-0}|=|{\frac{e^\frac{-1}{\sqrt{x^2+y^2}}}{e^{x^2+y^2}-1}-0}| = |{\frac{1}{e^{x^2+y^2}-1}*e^\frac{-1}{\sqrt{x^2+y^2}}}|$. Since ${e^\frac{-1}{\sqrt{x^2+y^2}}}<1$ for all $x,y\neq(0,0)$, we get $|{f(x,y)-0}|< |\frac{1}{e^{\delta^2}-1}*1|$. Setting $\delta=\sqrt{ln(\frac{1}{\epsilon}-1)}$ We can plug in to the inequality to get $|{f(x,y)-0}|<|\frac{1}{e^{\delta^2}-1}*1|<\epsilon$. Thus the limit is equal to $0$

Answer 1

$0<\sqrt {x^2 + y^2} = r < \delta$

$0<\frac {e^{\frac {-1}{r}}}{e^{r^2} + 1} < \frac 12 e^{\frac {-1}{r}}<\epsilon$

We can say this because $e^{r^2} > 1$ and $e^{r^2} + 1 > 2$

$-\frac {1}{r} < \ln 2\epsilon\\ r< -\frac {1}{\ln 2\epsilon}$

For any $\epsilon > 0,$ when $\delta < -\frac {1}{\ln 2\epsilon}$ then $|f(x,y)| < \epsilon$