I understand that an algebraic curve is genus 0 iff it can be parameterized using rational functions. I am curious if there is a simple way to know if a transcendental curve is genus 0? Specifically the curve: $$ 1-e^y-e^xe^{-y}=0 $$ Which can be parameterized as: $$ y=log(t+1/2), \hspace{10pt} x=log((t+1/2)(1/2-t)) $$
It seems to be genus 0, in that the $\:t\:$ parameter is defined on the Riemann sphere, except at $\:t=\pm \large\frac12$ (the natural logarithm singularity).
Is the above curve genus 0?