= Solution
Put the <stagnation point> at the origin and the local tangent plane at $z=0$. A regular <Taylor expansion> has no linear term because $\mathbf u(0)=0$. The impermeable tangent plane imposes $\partial_z\phi=0$ at $z=0$ to leading order, excluding mixed quadratic terms $xz,yz$. For the illustrated axisymmetric approach, <rotational symmetry> in the tangent plane makes the $x^2$ and $y^2$ coefficients equal and removes $xy$. Finally the <Laplace equation> fixes the normal coefficient to minus twice the tangential one. Up to an irrelevant additive constant,
$$
\boxed{\phi=W(x^2+y^2-2z^2)+O(|\mathbf r|^3),\qquad
\mathbf u=(2Wx,2Wy,-4Wz)+O(|\mathbf r|^2).}
$$
For incoming flow from $z>0$, $W>0$. This is a local leading form under the stated symmetry, not the exact universal potential near every possible wall <stagnation point>; a generic tangential strain can have unequal principal rates.
A <streamline> solves $\dot x=2Wx$, $\dot y=2Wy$, $\dot z=-4Wz$, hence $x=Ae^{2Wt}$, $y=Be^{2Wt}$, $z=Ce^{-4Wt}$. Eliminating the parameter gives
$$
\boxed{\arg(x+iy)=\text{constant},\qquad z(x^2+y^2)=\text{constant}.}
$$
Each <streamline> lies in a meridional plane. With cylindrical radius $s=\sqrt{x^2+y^2}$, its shape is $z=C_1/s^2$. The symmetry axis and wall are limiting invariant <streamlines>. A particle exactly on the incoming axis approaches the <stagnation point> asymptotically.
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