I know that if $F_n$ is the empirical distribution function and $F$ is the true distribution function, then, with $T$ being any statistical functional (satisfying the von-Mises derivative conditions), the following Central Limit Theorem holds:
$$\sqrt{n}(T(F_n)-T(F))\to N(0,E\varphi^2_F(X))-----(1)$$
in distribution, where $\int\varphi_F(x)dG=T'(G-F)=\dfrac{d}{d\alpha}T(\alpha(G-F)+F)|_{\alpha=0}$ for any two distribution functions $F$ and $G$.
Here, in deriving the above asymptotic result, our $G=F_n$, the standard empirical distribution function.
If we let $G=\Delta_y$, the degenerate distribution at $y$, then the influence function is $\varphi_F(y)=\dfrac{\dfrac{1}{2}-\Delta_y(x_0)}{f(x_0)}$ where $x_0=F^{-1}(\dfrac{1}{2})$. Here $f=F'$ is the correct pdf.
Do these immediately tell us that $$\sqrt{n}(\hat{\theta_n}-x_0)\to N\left(0,\dfrac{1}{4f^2(x_0)}\right)$$ Note that $E\varphi^2_F(X)=\dfrac{1}{4f^2(x_0)}$ but I simply cannot put this value in $(1)$ because my $G$ are different. There $G=F_n$ and here $G=\Delta_y$.
