jmc

The fraction $\frac{1}{2y+1}$ fails to be defined only if the denominator is zero. This occurs when $y$ is the solution of the equation $$2y+1=0,$$ which is $y=-\frac{1}{2}$. Thus the domain of $k(y)$ is $$\boxed{\left(-\infty ,-\frac{1}{2}\right)\cup \left(-\frac{1}{2},\infty \right)}.$$