Consider an electric field (or whatever 3-component vector field you want) $\mathbf{E}=\left(E_x, E_y, E_z\right)$. Let $\mathbf{r}(s) = (x(s), y(s), z(s))$ be the parametric equation of a field line, where $\mathrm{d} s$ is a line segment along the field line. Then, by definition, $$\frac{\mathrm{d} \mathbf{r}}{\mathrm{d} s} = \frac{\mathbf{E}(\mathbf{r})}{\lVert \mathbf{E}(\mathbf{r})\rVert}.$$
It follows that the field line (integral curve) equations are $$\frac{\mathrm{d} x}{E_x} = \frac{\mathrm{d} y}{E_y} = \frac{\mathrm{d} z}{E_z} = \frac{\mathrm{d} s}{\lVert\mathbf{E}\rVert}.$$
I now want to convert these into cylindrical coordinates but am lost on where to begin. I know how to convert the unit vectors: $$\begin{pmatrix} \hat{\boldsymbol{\rho}} \\ \hat{\boldsymbol{\phi}} \\ \hat{\mathbf{z}} \end{pmatrix} = \begin{pmatrix} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \hat{\mathbf{x}} \\ \hat{\mathbf{y}} \\ \hat{\mathbf{z}} \end{pmatrix} = \begin{pmatrix} \cos \phi\,\hat{\mathbf{x}} + \sin \phi\, \hat{\mathbf{y}} \\ -\sin \phi\, \hat{\mathbf{x}} + \cos \phi \hat{\mathbf{y}} \\ \hat{\mathbf{z}} \end{pmatrix}$$ and know how to convert the 'distances': $$x = \rho \cos \phi,\hspace{1em} y=\rho \sin \phi,\hspace{1em} z=z,$$ but don't know where to go from there.
Do I just substitute $x=\rho \cos \phi$ and so on into $\mathbf{r}$? Using the unit vectors, do I conclude that $E_{\rho} = E_x \cos \phi + E_y \sin \phi$ and similarly for $E_{\phi}$? And how do I put these together (mainly how I apply this to the original derivative expression)?
Any help would be appreciated.
The expression for your vector $(E_\rho, E_\phi,E_z)$ field is correct. For the derivative,you just need to use the total derivative definition, for example:
$$ dx = \frac{dx}{d\rho}d\rho +\frac{dx}{d\phi}d\phi +\frac{dx}{dz}dz, $$
or for a function $f(x,y,z)$ $$ \frac{df}{dx} = \frac{df}{d\rho}\frac{d\rho}{dx} + \frac{df}{d\phi}\frac{d\phi}{dx} + \frac{df}{dz}\frac{dz}{dx}. $$