Convert function from cartesian coordinates to cylindrical and spherical

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Introduction

I have a point charge of magnitude $-6Q_0$. This point charge is placed in the origin of an orthogonal coordinate system. The electric field at a arbitrary point $P=(x,y,z)$ caused by this point charge can be expressed by the function $$\mathbf{E}(x,y,z)=\mathbf{a_x}\Bigg(\frac{k_c \cdot-6Q_0 \cdot x}{(x^2+y^2+z^2)^{\frac{3}{2}}} \Bigg) + \mathbf{a_y}\Bigg(\frac{k_c \cdot-6Q_0 \cdot y}{(x^2+y^2+z^2)^{\frac{3}{2}}} \Bigg) + \mathbf{a_z}\Bigg(\frac{k_c \cdot-6Q_0 \cdot z}{(x^2+y^2+z^2)^{\frac{3}{2}}} \Bigg)$$

Where $\mathbf{E}(x,y,z)$ is the electric field at a point $P=(x,y,z)$ and $\mathbf{a_x,a_y,a_z}$ are unit vectors pointing towards the x-axis, y-axis and z-axis respectively. $k_c$ is just a constant.

Note: The expression for the above electric field is derived through Coulomb's law: $$\mathbf{E}=k_c\cdot\frac{Q}{|r|^2}\cdot\frac{\mathbf{r}}{|r|} $$

Problem

How can I express the electric field at a given point $P$ in terms of cylindrical and spherical coordinates.

That is, how do I convert my expression from cartesian coordinates to cylindrical and spherical so that the expression for the electric field looks like this for the cylindrical:

$$\mathbf{E}(r,\phi,z) $$

And like this for the spherical coordinatsystem:

$$\mathbf{E}(R,\theta,\phi) $$

Is there some method to convert an entire expression into a different coordinate system? I hope someone can help me with this.

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On BEST ANSWER

Let $C= -6 Q_0 k_c$. Using standard resolution of vectors

Cylindrical

$$ E(r,\phi,z) =\dfrac{ C(r \cos \phi, r \sin\phi,z)}{(r^2+z^2)^{\frac32}} $$

Spherical

$$ {E(R,\phi,\theta)=\dfrac{ C(\cos \phi \cos \theta, \cos \phi \sin \theta,\sin\phi)} {R^2} } $$

Or in the form you have written in your comment:

$$\mathbf{E}(r,\phi,z)=\mathbf{a_r}(\frac{C \cdot r\cos(\phi)}{(r^2+z^2)^{\frac{3}{2}}})+\mathbf{a_{\phi}}(\frac{C \cdot r\sin(\phi)}{(r^2+z^2)^{\frac{3}{2}}})+\mathbf{a_z}(\frac{C \cdot z}{(r^2+z^2)^{\frac{3}{2}}})$$