Do we have
$\int_U |Du|^2 dx + \int_{\partial U}u^2 dx < \infty \implies \int_U |u|^2 dx < \infty$ when $u \in H^1(U)$ and $\partial U $ is smooth ?
This result is true for $u \in H_0^1(U)$ by Poincaré's inequality; complicating the boundary conditions seems to make this question harder.
The following stronger claim is valid: There is $c>0$ such that $$ \|u\|_{L^2(\Omega)} \le c ( \|Du\|_{L^2(\Omega)} + \|u\|_{L^2(\partial \Omega)}). $$ Suppose not, than for all $n$ there is $u_n$ such that $$ \|u_n\|_{L^2(\Omega)} > n ( \|Du_n\|_{L^2(\Omega)} + \|u_n\|_{L^2(\partial \Omega)}). $$ Then $u\ne0$, and we can scale the inequality to $\|u_n\|_{L^2(\Omega)}=1$. So $(u_n)$ is a bounded sequence in $H^1(\Omega)$. And there is a subsequence $(u_{n'})$ converging weakly in $H^1(\Omega)$ and strongly in $L^2(\Omega)$ to some $u$. Due to this strong convergence, $\|u\|_{L^2(\Omega)}=1$.
In addition, the inequality implies $Du_n\to0$ in $L^2(\Omega)$ and $u_n\to 0$ in $L^2(\partial\Omega)$. Hence $u$ is a constant function (because of $Du=0$) with zero boundary values. Hence $u=0$. Which is a contradiction.