Suppose $u \in H^1_{\text{loc}}(\mathbb{R^2})$. I want to show that if $\Delta u = |\nabla u|^{3/2}$ in the distributional sense then $u \in C^{\infty}$.
I know that since $\nabla u \in L^2$ (I'll omit the $_{\text{loc}}$ subscripts) then $|\nabla u|^{3/2} \in L^{4/3}$ hence by standard regularity results for the Laplacian we get $u \in W^{2,4/3}$ then Sobolev embedding gives $u \in W^{1,4}$, so $|\nabla u |^{3/2} \in L^{8/3}$ hence (Laplacian regularity) $u \in W^{2,8/3}$ and so $u \in C^1$.
If $u \in C^1$ then $|\nabla u|^{4/3}$ is continuous, so $u \in C^2$, then $|\nabla u|^{4/3}$ is $C^1$ and $u$ is $C^3$, but I don't know how to go further. $|x|^{4/3}$ is a $C^1$ function, but not more than that so I don't know how to continue my bootstrapping. If the right hand side was some smooth function of $\nabla u$ then it would be fairly simple, but it's not the case here - I'd be grateful for suggestions on how to deal with that.