If Delta u=0 , show u_x ; xu_x+yu_y+zu_z ; yu_x-xu_y ; and…
If $\Delta u=0$, show $u_x$; $xu_x+yu_y+zu_z$; $yu_x-xu_y$; and $y^2u_{xx}-2xyu_{xy}+x^2u_{yy}+zu_z$ are all harmonic.
Explanation of Method
For the constant-coefficient case ($u_x$) use that $\Delta$ commutes with $\partial_x$. For the variable-coefficient combinations, compute $\Delta$ of each term directly via the product rule, and note the "pure third/fourth order" pieces always regroup into $\partial^\beta(\Delta u)=0$, leaving only lower-order cross terms that cancel between the pieces of the combination.
Step 1 · (i) $u_x$
$\Delta(u_x)=\partial_x(\Delta u)=0$ since $\Delta$ has constant coefficients. \checkmark (Likewise $u_y,u_z$.)
Step 2 · (ii) $xu_x+yu_y+zu_z$
Direct computation gives $\Delta(xu_x)=2u_{xx}+x\,\partial_x(\Delta u)=2u_{xx}$, and cyclically $\Delta(yu_y)=2u_{yy}$, $\Delta(zu_z)=2u_{zz}$; summing, $\Delta(xu_x+yu_y+zu_z)=2\Delta u=0$.
Step 3 · (iii) $yu_x-xu_y$
$\Delta(yu_x)=2u_{xy}+y\,\partial_x(\Delta u)=2u_{xy}$ and $\Delta(xu_y)=2u_{xy}$ (by symmetry), so $\Delta(yu_x-xu_y)=2u_{xy}-2u_{xy}=0$.
Step 4 · (iv) $y^2u_{xx}-2xyu_{xy}+x^2u_{yy}+zu_z$
Term by term: $\Delta(y^2u_{xx})=2u_{xx}+4yu_{xxy}+y^2\partial_x^2(\Delta u)$, $\Delta(xyu_{xy})=2yu_{xxy}+2xu_{xyy}+xy\,\partial_x\partial_y(\Delta u)$, $\Delta(x^2u_{yy})=2u_{yy}+4xu_{xyy}+x^2\partial_y^2(\Delta u)$, $\Delta(zu_z)=2u_{zz}$. All the $\partial(\Delta u)$ terms vanish, and the combination $[2u_{xx}+4yu_{xxy}]-2[2yu_{xxy}+2xu_{xyy}]+[2u_{yy}+4xu_{xyy}]+2u_{zz}$ has the $4yu_{xxy}$ and $4xu_{xyy}$ terms cancel exactly, leaving $2u_{xx}+2u_{yy}+2u_{zz}=2\Delta u=0$.
Verification
All four identities were checked symbolically against several concrete harmonic test functions (e.g. $u=xz$, $u=x^2-y^2$, $u=x^3-3xy^2$), confirming $\Delta v=0$ in every case. \checkmark
Result
Conclusion: all four listed functions are harmonic whenever $u$ is.