If I understand things correctly, $u$ is vanishing on some non-empty open subset. Moreover, you have pointwisely on $\mathbb R^3$ the differential inequality $$ \vert \Delta u\vert\le C\vert u\vert^5. $$ Also, you know that the function $u$ is in $\dot H^1(\mathbb R^3)$, which is a subset of $L^6(\mathbb R^3)$. This implies that $$ \vert u\vert^5=\vert u\vert\underbrace{\vert u\vert^4}_{L^{3/2}}, $$ implying that you have a differential inequality $ \vert \Delta u\vert\le V\vert u\vert, \ V \in L^{d/2} $ in $\mathbb R^d, d\ge 3$. Then the strong unique continuation result due to Jerison & Kenig [MR0794370] entails that $u\equiv 0$. Note that $d/2$ is critical and that $L^{d/2}_{\text{loc}}$ is enough.

If I understand things correctly, $u$ is vanishing on some non-empty open subset. Moreover, you have pointwisely on $\mathbb R^3$ the differential inequality $$ \lvert \Delta u\rvert\le C\lvert u\rvert^5. $$ Also, you know that the function $u$ is in $\smash{\dot H}^1(\mathbb R^3)$, which is a subset of $L^6(\mathbb R^3)$. This implies that $$ \lvert u\rvert^5=\lvert u\rvert\underbrace{\lvert u\rvert^4}_{L^{3/2}}, $$ implying that you have a differential inequality $ \lvert \Delta u\rvert\le V\lvert u\rvert, \ V \in L^{d/2} $ in $\mathbb R^d$, $d\ge 3$. Then the strong unique continuation result due to Jerison & Kenig [MR0794370] entails that $u\equiv 0$. Note that $d/2$ is critical and that $L^{d/2}_{\text{loc}}$ is enough.