https://en.wikipedia.org/wiki/Helmholtz%27s_theorems
In fluid dynamics, according to Helmholtz's Thm 1, the strength of a vortex filament is constant along its length. A vortex filament is the streamline of the vector field curl(v), where v is the fluid velocity.
Let v=c(z)*[-y,x,0]. The vorticity w=curl(v)=[-x c'(z), y c'(z), 2 c(z)]. The vorticity has one streamline running vertically at x=y=0 with varying magnitude dependent on c(z). This seems to be a counterexample to Helmholtz's Thm 1.
Am I applying the Thm incorrectly somehow? or misinterpreting something?
A vortex filament is a vortex tube of infinitesimal cross-sectional area. This is a structure that is aligned with the direction of vorticity and either the vorticity field vanishes or is parallel to the filament throughout a deleted neighborhood. In this case Helmholtz's first theorem applies in that the strength (the flux of vorticity through a disk perpendicular to the filament as the radius of the disk shrinks to zero) does not vary along the filament.
For example, imagine that we have a vortex filament aligned with the $z-$axis. Consider a tube with circular cross section of radius $R$ and length $Z$ enclosing the region
$$\Omega =\left\{x,y,z ) : \,\,0 \leqslant \sqrt{x^2 + y^2} \leqslant R, \,\,0 \leqslant z \leqslant Z \right\}$$
The boundary $\partial \Omega = S_0 \cup S_R \cup S_Z$ where $S_0$ and $S_Z$ are circular disks of radius $R$ located in the planes $z = 0$ and $z = Z$, respectively, and $S_R$ is the lateral surface
$$S_R = \left\{(x,y,z) : \,\, \sqrt{x^2 + y^2} = R, \,\, 0 \leqslant z \leqslant Z \right\}$$
Since vorticity has zero divergence, $\nabla \cdot \mathbf{\omega} = 0$, we have by the divergence theorem,
$$\tag{*} \int_{S_0} \mathbf{\omega} \cdot \mathbf{n} \, dS + \int_{S_Z} \mathbf{\omega} \cdot \mathbf{n} \, dS + \int_{S_R} \mathbf{\omega} \cdot \mathbf{n} \, dS = \oint_{\partial \Omega} \mathbf{\omega} \cdot \mathbf{n} \, dS = \int_{\Omega}\nabla \cdot \mathbf{\omega} \, dV = 0$$
With a vortex filament, we have either $\mathbf{\omega} = 0$ or $\mathbf{\omega} \cdot \mathbf{n} = 0$ on $S_R$. This implies
$$-\int_{S_0} \mathbf{\omega} \cdot \mathbf{n} \, dS = \int_{S_Z} \mathbf{\omega} \cdot \mathbf{n} \, dS,$$
and since $\mathbf{n} = -\mathbf{e_z}$ on $S_0$ and $\mathbf{n} = \mathbf{e_z}$ on $S_z$ we see that the strength of the filament does not depend upon $z$.
With your velocity field, the $z-$axis or any parallel line is not a vortex filament and there is no counterexample.
Note that the correct vorticity for this flow is $(\omega_x,\omega_y, \omega_z) = (-xc'(z), -yc'(z), 2c(z))$. The integrals over the top and bottom surfaces, $S_0$ and $S_Z$, are
$$\int_{S_0} \mathbf{\omega} \cdot \mathbf{n} \, dS = - \int_{0}^{R}\int_{0}^{2\pi} \left.\omega_z\right|_{z = 0} R\, dr\, d\theta = -2\pi R^2c(0), \\ \int_{S_Z} \mathbf{\omega} \cdot \mathbf{n} \, dS = \int_{0}^{R}\int_{0}^{2\pi} \left.\omega_z\right|_{z = Z} R\, dr\, d\theta = 2\pi R^2 c(Z),$$
and the lateral surface integral is
$$\begin{align}\int_{S_R} \mathbf{\omega} \cdot \mathbf{n} \, dS &= \int_{0}^{2\pi}\int_0^Z (\omega_x n_x + \omega_y n_y)\,R\, d\theta \, dz \\&= \int_{0}^{2\pi}\int_0^Z c'(z)\left.\left( -x \frac{x}{\sqrt{x^2 + y^2}} - y\frac{y}{\sqrt{x^2 + y^2}}\right)\right|_{\sqrt{x^2+y^2} = R}\,R\, d\theta \, dz \\ &= -\int_{0}^{2\pi}\int_0^Z c'(z) \,R^2\, d\theta \, dz \\ &= -2 \pi R^2 \int_0^Z c'(z) \, dz \\ &= -2\pi R^2 (c(Z) - c(0)) \end{align}.$$
If you ignore the contribution from the lateral surface integral, you erroneously conclude that
$$0 = \int_{S_0} \mathbf{\omega} \cdot \mathbf{n} \, dS + \int_{S_Z} \mathbf{\omega} \cdot \mathbf{n} \, dS = 2\pi R^2 c(Z) - 2\pi R^2 c(0),$$
which implies that we must have $c(Z) = c(0)$ to ensure that the vortex tube strength is constant along its length, suggesting a counterexample to Helmholtz's first theorem.
However, the contribution from the lateral surface integral always cancels the contributions from the ends so that (*) holds for any $Z$ and there is no contradiction when $c(z)$ is not constant.