A harmonic function degenerate in one direction

Question. Let $$u: B^3 \to \mathbf{R}$$ be a harmonic function with $$u(0) = 0$$, $$Du(0) = 0$$, where its homogeneous harmonic blow-up is a polynomial $$p = p(x,y)$$ in two variables, so independent of $$z$$; in other words $$p$$ is a non-zero homogeneous harmonic polynomial so that $$$$u(x,y,z) = p(x,y) + o( \lvert (x,y,z) \rvert^m),$$$$ where $$2 \leq m = \operatorname{deg} p$$. Must $$u$$ be translation-invariant with respect to $$z$$? Can the origin be isolated in the singular set $$u^{-1}(0) \cap \lvert Du \rvert^{-1}(0)$$?

• Could you define the homogenuous harmonic blow-up? 2 days ago
• The answer to the first question is negative: a counterexample is $u(x,y,x)=P_m(x,y)+Q_n(x,y,z),$ where $P_m,P_n$ are homogeneous harmonic polynomials of degrees $n>m$, and $P_n$ depends on all three variables. 2 days ago
• @AlexandreEremenko Of course, how embarrassing that I missed that - thanks for pointing this out! 2 days ago
• @AlexandreEremenko By the way, if you wanted to post your comments as an answer, I'd be happy to accept it! yesterday
• @Leo Moos: My second comment was not correct, and I deleted it. Anyway, you solved the problem yourself, probably inspired by the first (trivial) comment. yesterday

The questions have been answered in the comments, I am just recording them here: Alexandre Eremenko pointed out that no, the function $$u$$ need not be translation-invariant, because the dependencies on $$z$$ could be 'hidden' inside a polynomial of higher degree, say $$$$u = p(x,y) + q(x,y,z),$$$$ with $$\operatorname{deg} q > \operatorname{deg} p$$.
This also gives a hint for the second question: the answer is yes, there exist examples of such $$u$$ that only have isolated singularity at the origin. The example given below $$u$$ is basically of the form above—with $$q$$ picked so as to have an isolated singularity at the origin—, except for the fact that one multiplies $$q$$ by a small constant to avoid introducing new singular points.
Specifically, pick a constant $$\delta \in (0,1/3)$$ and define $$$$u(x,y,z) = x^2 - y^2 + \delta(2x^3 - 3xy^2 - 3xz^2).$$$$ Then $$$$Du(x,y,z) = (2x + \delta( 6x^2 - 3y^2 - 3z^2),-2y - 6\delta xy,-6\delta xz).$$$$ At a critical point $$(x,y,z)$$:
• from $$D_y u = 0$$ one finds that $$y(1 + 3\delta x) = 0$$. As $$\delta < 3$$, the second factor never vanishes if $$\lvert x \rvert < 1$$, so $$y = 0$$;
• from $$D_z u = 0$$ one finds that either $$x = 0$$ or $$z = 0$$.
• from $$D_x u = 0$$, if $$x = 0$$ then immediately $$z = 0$$. If instead $$z = 0$$ then $$0 = D_x u = 2x + 6\delta x^2 = 2x(1 + 3 \delta x)$$. Again, our choice of a sufficiently small $$\delta$$ means that $$1 + 3 \delta x > 0$$ on $$B^3$$, so $$x = 0$$.
Therefore $$Du(x,y,z) = 0$$ is equivalent to $$(x,y,z) = 0$$. Obviously $$u(0,0,0) = 0$$, so this is indeed the unique singular point.