$\newcommand{\na}{\nabla}\newcommand{\R}{\mathbb R}$Without loss of generality, $C=1$, so that 
\begin{equation}
	f(x)\equiv\exp\Big\{-\frac{\pi |x|^2}{2a^2}\Big\}. 
\end{equation}
Hence, $\na f(x)=-f(x)\pi x/a^2$, $|\na f(x)|^2=f(x)^2\pi^2|x|^2/a^4$, and the left-hand side of the desired equality is (with $\|\cdot\|_2:=\|\cdot\|_{L^2}$ and $x=(x_1,\dots,x_n)\in\R^n$), 
\begin{equation}
\begin{aligned}
	lhs&=\frac{a^2}\pi\|\na f(x)\|_2^2 \\ 
	&=\frac{\pi}{a^2}\int_{\R^n} dx\,f(x)^2|x|^2 \\ 
	&=\frac{\pi}{a^2}\int_{\R^n} dx\,f(x)^2\sum_1^n x_i^2 \\ 
	&=n\,\frac{\pi}{a^2}\int_{\R^n} dx\,f(x)^2 x_1^2 \\ 
		&=n\,\frac{\pi}{a^2}\int_{\R^n} f(x)^2 x_1^2\,\prod_1^n dx_i \\ 
		&=n\,\frac{\pi}{a^2}\int_{\R}dx_1\,\exp\Big\{-\frac{\pi x_1^2}{a^2}\Big\} x_1^2\  
		\prod_2^n \int_{\R} dx_i\,\exp\Big\{-\frac{\pi x_i^2}{a^2}\Big\} \\ 
		&=n\,\frac{\pi}{a^2}\frac{a^3}{2\pi}\,a^{n-1}=\frac n2\,a^n.
\end{aligned}
\end{equation}
Similarly, 
\begin{equation}
\begin{aligned}
\|f\|_2^2  
	&=
		\prod_1^n \int_{\R} dx_i\,\exp\Big\{-\frac{\pi x_i^2}{a^2}\Big\}=a^n.
\end{aligned}
\end{equation}
So, the right-hand side of the desired equality is 
\begin{equation}
\begin{aligned}
	rhs&=-\frac{\pi}{a^2}\int_{\R^n} dx\,f(x)^2|x|^2
	-\|f\|_2^2 \ln(\|f\|_2^2)
	+n(1+\ln a)\|f\|_2^2 \\ 
	&=-lhs-\|f\|_2^2 \ln(\|f\|_2^2)+n(1+\ln a)\|f\|_2^2 \\
	&=-\frac n2\,a^n-a^n \ln(a^n)+n(1+\ln a)a^n=\frac n2\,a^n=lhs,
\end{aligned}
\end{equation}
as desired.