Three-point function in CFT Reason for deriving the 3-point function
Searching for a derivation of the three-point function constraints in CFT online I have realised that there is no derivation of the 3-point function. Most authors derive the 2-point function and leave the other one as an exercise to the reader.
In trying to derive the 3-point function I realised that there are some subtleties I don't fully understand.
Derivation:
We define the 3-point correlation function as$$
\langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\mathcal{O}_3(x_3)\rangle=f(x_1,x_2,x_3)
$$
Translational symmetry
We impose the condition that $$
\left(\frac{\partial}{\partial x_1}+\frac{\partial}{\partial x_2}+\frac{\partial}{\partial x_3}\right)f(x_1,x_2,x_3)=0
$$
From the derivtion of the 2-point function I understand that this gives $$
f=f(X_{12}^\mu X_{13}^\mu X_{23}^\mu)
$$
where $X_{ij}=(x_i-x_j)^\mu$, but can we show how this is the case from the translational symmetry condition imposed above?
Lorentz group symmetry
Here if we assume the above we have
$$
f=f(X_{12}^2 X_{13}^2 X_{23}^2)
$$
Dilatation symmetry
This is where I'm not quite sure how to proceed. We should satisfy the following (correct me if I'm wrong)
$$
\left(x_1^\mu \frac{\partial}{\partial x_1^\mu}+x_2^\mu \frac{\partial}{\partial x_2^\mu}+x_3^\mu \frac{\partial}{\partial x_3^\mu}+\Delta_1+\Delta_2+\Delta_3\right)f(X_{12}^2 X_{13}^2 X_{23}^2)=0
$$
Not sure how I can solve this equation.
Special Conformal symmetry
Finally we must satisfy
$$
(-2x_{1\mu}\Delta_1-2x_{2\mu}\Delta_2-2x_{3\mu}\Delta_3+k_{1\mu}+k_{2\mu}+k_{3\mu})f(X_{12}^2 X_{13}^2 X_{23}^2)=0
$$
with $$k_{i\mu}=x^2\frac{\partial}{\partial x_{i\mu}}-2x_\mu x^\nu\frac{\partial}{\partial x_{i\nu}}$$
Which again, I'm not sure how to calculate.
Disclaimer
I think it would be useful to have this derivation available with all the steps carefully shown so I'm hoping that with several hints I can complete this.
 A: You're almost there!
Remembering how dilatations act on $\mathcal{O}_i(x_i)$, to demand covariance of the 3-pt funstion you need$^{(*)}$
$$\big\langle \mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\mathcal{O}_3(x_3)\big\rangle = \frac{C_{123}}{x_{12}^a\,x_{23}^b x_{31}^c}, $$
where $x_{ij}:=|x_i-x_j|$ (I change your $X_{ij}$ notation to not carry an extra square around) and
$$a+b+c = \Delta_1+\Delta_2+\Delta_3 \label{1}\tag{1}.$$ Now, under SCT, if $y_i := 1 - 2 b_\mu x^\mu_i + b^2 x^2_i$, you need the transformed 3-pt function to equal the untransformed, i.e.
$$ \frac{C_{123}}{x_{12}^a\,x_{23}^b x_{13}^c} = \frac{C_{123}}{y_{1}^{\Delta_1}\,y_{2}^{\Delta_2} y_{3}^{\Delta_3}}\;\frac{(y_1 y_2)^{a/2}(y_2 y_3)^{b/2}(y_3 y_1)^{c/2}}{x_{12}^a\,x_{23}^b x_{31}^c}, $$
with
$$ a+c= 2\Delta_1, \qquad a+b = 2\Delta_3, \qquad b+c = 2\Delta_3 \label{2}\tag{2}$$
From here on \eqref{1} and \eqref{2} have unique solution 
$$ a = \Delta_1 + \Delta_2 - \Delta_3$$
$$ b = \Delta_2 + \Delta_3 - \Delta_1$$
$$ c = \Delta_3 + \Delta_1 - \Delta_2,$$
which gives you in turn the 3-pt function you know and love
$$\big\langle \mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\mathcal{O}_3(x_3)\big\rangle = \frac{C_{123}}{x_{12}^{\Delta_1 + \Delta_2 - \Delta_3}\,x_{23}^{\Delta_2 + \Delta_3 - \Delta_1} x_{31}^{\Delta_3 + \Delta_1 - \Delta_2}}.$$
Edit: On translations and Lorentz transformations
For translations we have that $x_i \mapsto x_i + s$. Then $x_i-x_j$ is invariant, so $f(x_1,x_2,x_3)$ needs to be $f(x_1-x_2,x_2-x_3,x_3-x_1)$. Now from Lorentz you get immediately that they should only come in the form $|x_i-x_j|=x_{ij}$, i.e. $f=f(x_{12},x_{23},x_{31})$ (Lorentz is like rotations!). The fact that it is multiplicative really comes from the dilatation covariance (cf. footnote).
A general comment
It is far more useful to demand covariance of an $n$-point function based on the transformation of the operators themselves and not via those differential equations. Namely, covariance of an $n$-pt function means that under any conformal transformation $x\mapsto x'$:
$$\big\langle \mathcal{O}_1(x_1)\mathcal{O}_2(x_2) \cdots \mathcal{O}_n(x_n)\big\rangle \overset{!}{=} \left\vert\frac{\partial x'}{\partial{x}}\right\vert^{\Delta_1/d}_{x=x_1} \left\vert\frac{\partial x'}{\partial{x}}\right\vert^{\Delta_2/d}_{x=x_2} \kern-3mm\cdots \left\vert\frac{\partial x'}{\partial{x}}\right\vert^{\Delta_n/d}_{x=x_n} \big\langle \mathcal{O}_1(x_1')\mathcal{O}_2(x_2') \cdots \mathcal{O}_n(x_n')\big\rangle.$$

$^{(*)}$ In fact you should in principle have
$$\big\langle \mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\mathcal{O}_3(x_3)\big\rangle = \sum_{\begin{array}{c} a,b,c \\ a+b+c = \Delta_1+\Delta_2+\Delta_3 \end{array}}\frac{C_{123}}{x_{12}^a\,x_{23}^b x_{31}^c}, $$
but there are unique $a,b,c$ (see main argument), thus only one term in the sum.
