It is not true that $\dot\rho(\phi_1,\phi_1)=0$: the object $\rho(\phi_1,\phi_1)$ does not exist.

It doesn't make sense to talk about $\rho$ for $\phi_1=\phi_2$. The object $\rho$ is a distribution (as a function of $\phi$, not $x$) so it is meaningless to "evaluate" it on a given function.

Indeed, $$ \langle\phi_1|\phi_2\rangle=\delta(\phi_1-\phi_2) $$ where $\delta$ is the Dirac delta in the space of functionals (i.e., with respect to functional integration, not regular integration). See this PSE post for some (formal) applications of this formula.

The density matrix $\rho(\phi_1,\phi_2)$ only makes sense under the integral sign: $$ \int_{\mathcal F^2}\rho(\phi_1,\phi_2)F[\phi_1,\phi_2]\,\mathrm d\phi_1\mathrm d\phi_1 $$ where $\mathcal F$ denotes some reasonable function space and $F$ is a healthy functional on it.

(This would be equivalent to asking about $\rho(x,x')$ at $x=x'$. The ket $|x\rangle$ depends on $x$ as a distribution, not as a regular function. So it is meaningless to ask what happens at $x=x'$; distributions do not depend on integration variables. In QFT the situation is even more singular because $|\phi\rangle$ is an eigenket of an operator which is itself a distrubution.)

It is not true that $\dot\rho(\phi_1,\phi_1)=0$: the object $\rho(\phi_1,\phi_1)$ does not exist.

It doesn't make sense to talk about $\rho$ for $\phi_1=\phi_2$. The object $\rho$ is a distribution (as a function of $\phi$, not $x$) so it is meaningless to "evaluate" it on a given function.

Indeed, $$ \langle\phi_1|\phi_2\rangle=\delta(\phi_1-\phi_2) $$ where $\delta$ is the Dirac delta in the space of functionals (i.e., with respect to functional integration, not regular integration). See this PSE post for some (formal) applications of this formula.

The density matrix $\rho(\phi_1,\phi_2)$ only makes sense under the integral sign: $$ \int_{\mathcal F^2}\rho(\phi_1,\phi_2)F[\phi_1,\phi_2]\,\mathrm d\phi_1\mathrm d\phi_1 $$ where $\mathcal F$ denotes some reasonable function space and $F$ is a healthy functional on it.