The interval contained in the domain of definition of non-zero solutions of the differential equation

$(x-3{)}^2{y}^{\mathrm{\prime }}+y=0,$ is

We have,

$\begin{array}{l} (x-3{)}^2{y}^{\mathrm{\prime }}+y=0\\ \Rightarrow (x-3{)}^2\frac{dy}{dx}+y=0\Rightarrow \frac{1}{y}dy+\frac{1}{(x-3{)}^2}dx=0\end{array}$

Integrating, we get

$\begin{array}{l} \log \left|y\right|-\frac{1}{x-3}=\log C\\ \Rightarrow \log \left\{\frac{\left|y\right|}{C}\right\}=\frac{1}{x-3}\Rightarrow \left|y\right|=C{e}^{-\frac{1}{x-3}}\end{array}$

The domain of definition of this solution is $R-\left\{3\right\}.$

Clearly, $(-\pi /2,\pi /2)\subset R-\left\{3\right\}$. So, option (a) is correct.