Discovering vector potential from moving field analysis First I would like to remind Lorentz transformation of length and time as a matrix:
$$ \begin{pmatrix} 
ct'\\
x'\\ 
y'\\
z' 
\end{pmatrix} = \begin{pmatrix}
\gamma & -  \frac{v}{c} \gamma & 0 & 0 \\
- \frac{v}{c} \gamma & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} \cdot \begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}  $$
The Transform matrix 4x4 should be the same for all four vectors in Special Relativity.
Look at the picture below:

We know only scalar potential in both cases. My goal is to find vector potentials and to prove that
$$ \vec{E} = - \vec{\nabla} \phi - \frac{\partial \vec{A}}{\partial t} $$
Let me use Lorentz transformation for some imaginery four electric vector:
$$ \begin{pmatrix} 
\lambda \phi'\\
A_x'\\ 
A_y'\\
A_z' 
\end{pmatrix} = \begin{pmatrix}
\gamma & -  \frac{v}{c} \gamma & 0 & 0 \\
- \frac{v}{c} \gamma & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} \cdot \begin{pmatrix} \lambda \phi \\ A_x \\ A_y \\ A_z \end{pmatrix}  $$
$$ \phi '= - \frac{Q}{4 \pi \epsilon_0 r'}  $$
and
$$ \phi = - \frac{1}{4 \pi \epsilon_0 } \cdot \frac{Q}{r' - \vec{r'} \cdot \frac{\vec{v}}{c} } $$
$\lambda$ is some constant for a while
It is looking even good until I begin manipulate the algebra to get $ A_x' $
What do You think about this approach?
Maybe Am I totally wrong?
Please help
 A: If I understood well you are trying to deduce an expression for $\boldsymbol{A}$. I'll try to give it just by a retarded-time discussion without explicity passing from Lorentz transformations. You should consider a particle with charge $q_i$ moving along a trajectory $\boldsymbol{r}_i(t)$ with a velocity $\dot{\boldsymbol{r}}_i(\tau)\doteq\text{d}\boldsymbol{r}_i(\tau)/\text{d}\tau$ where $\tau$ will be the retarded time, defined at the end of the argumentation. Let's start from the equations of the Lorentz gauge
\begin{gather*}
\frac{1}{\mu}\nabla\cdot\boldsymbol{A}
+\varepsilon\frac{\partial\phi}{\partial t}=0
\\
\nabla^2\phi
-\mu\varepsilon\frac{\partial^2\phi}{\partial t^2}
=
-\frac{\rho}{\varepsilon}
\\
\nabla^2\boldsymbol{A}
-\mu\varepsilon\frac{\partial^2 \boldsymbol{A}}{\partial t^2}
=
-\mu\boldsymbol{j}
\end{gather*}
which have solutions
\begin{gather*}
\phi(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\int\limits_V
\frac{
\rho\left(\boldsymbol{r}^\prime,t
-\displaystyle{
\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}
}\right)
}{
|\boldsymbol{r}-\boldsymbol{r}^\prime|
}
\text{d}^3{\boldsymbol{r}^\prime}
\\
\boldsymbol{A}(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\int\limits_V
\frac{
\boldsymbol{j}\left(\boldsymbol{r}^\prime,t
-\displaystyle{
\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}
}\right)
}{
|\boldsymbol{r}-\boldsymbol{r}^\prime|
}
\text{d}^3{\boldsymbol{r}^\prime}
\end{gather*}
but you know what $\rho,\boldsymbol{j}$ are for a moving particle
\begin{gather*}
\rho_i(\boldsymbol{r},t)
=
q_i\delta(\boldsymbol{r}-\boldsymbol{r}_i(t))
\\
\boldsymbol{j}_i(\boldsymbol{r},t)
=
q_i\boldsymbol{u}_i(t)\delta(\boldsymbol{r}-\boldsymbol{r}_i(t))
\\
\phi_i(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\int\limits_V
\frac{q_i\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i\left(t
-\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}\right)\right)
}{
|\boldsymbol{r}-\boldsymbol{r}^\prime|
}
\text{d}^3{\boldsymbol{r}^\prime}
\\
\boldsymbol{A}_i(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\int\limits_V
\frac{q_i\dot{\boldsymbol{r}}_i\left(t-\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}\right)\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i\left(t
-\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}\right)\right)
}
{|\boldsymbol{r}-\boldsymbol{r}^\prime|
}
\text{d}^3{\boldsymbol{r}^\prime}
\end{gather*}
But you don't like very much this integral so you operate a substitution
\begin{gather*}
t^\prime
\doteq
t
-\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}
\\
\boldsymbol{d}_i(t)
\doteq
\boldsymbol{r}-\boldsymbol{r}_i(t)
\end{gather*}
such that the condition posed by the $\delta$ becomes $\boldsymbol{r}^\prime=\boldsymbol{r}-\boldsymbol{d}_i(t^\prime)$. So now you have
\begin{gather*}
\phi_i(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\int\limits_V
\frac{q_i\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^\prime)\right)}{|\boldsymbol{d}_i(t^\prime)|}
\text{d}^3{\boldsymbol{r}^\prime}
\equiv
\frac{1}{4\pi\epsilon}
\int\limits_{t^{\prime\prime}}
\int\limits_V
\frac{q_i\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^{\prime\prime})\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|}
\delta(t^{\prime\prime}-t^\prime)
\text{d}^3{\boldsymbol{r}^\prime}\text{d}{t^{\prime\prime}}
\\
\boldsymbol{A}_i(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\int\limits_V
\frac{q_i\dot{\boldsymbol{r}}_i(t^\prime)\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^\prime)\right)}{|\boldsymbol{d}_i(t^\prime)|}
\text{d}^3{\boldsymbol{r}^\prime}
\equiv
\frac{\mu}{4\pi}
\int\limits_{t^{\prime\prime}}\int\limits_V
\frac{q_i\dot{\boldsymbol{r}}_i(t^{\prime\prime})\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^{\prime\prime})\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|}
\delta(t^{\prime\prime}-t^\prime)
\text{d}^3{\boldsymbol{r}^\prime}\text{d}{t^{\prime\prime}}
\\
\phi_i(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\int\limits_{t^{\prime\prime}}
\frac{q_i\delta\left(t^{\prime\prime}-t+\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}_i(t^{\prime\prime})|}{c}}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|}
\text{d}{t^{\prime\prime}}
\\
\boldsymbol{A}_i(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\int\limits_{t^{\prime\prime}}
\frac{q_i\dot{\boldsymbol{r}}_i(t^{\prime\prime})\delta\left(t^{\prime\prime}-t+\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}_i(t^{\prime\prime})|}{c}}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|}
\text{d}{t^{\prime\prime}}
\end{gather*}
You see? Now the integral is on the time domain! We are near the end. Now a new variable substitution
\begin{equation*}
t^{\prime\prime\prime}
=
t^{\prime\prime}
-t
+\frac{|\boldsymbol{d}_i(t^{\prime\prime})|}{c}
\Longrightarrow
\text{d}{t^{\prime\prime\prime}}
=
\text{d}{t^{\prime\prime}}
+\frac{1}{c}\frac{\text{d}|\boldsymbol{d}_i(t^{\prime\prime})|}{\text{d}t^{\prime\prime}}\text{d}{t^{\prime\prime}}
\end{equation*}
but defining
\begin{equation*}
\boldsymbol{n}_i(t^{\prime\prime})
\doteq
\frac{\boldsymbol{d}_i(t^{\prime\prime})}{|\boldsymbol{d}_i(t^{\prime\prime})|}
\end{equation*}
You will see that
\begin{equation*}
{\text{d}|\boldsymbol{d}_i(t^{\prime\prime})|}{\text{d}t^{\prime\prime}}=-\boldsymbol{n}_i(t^{\prime\prime})\cdot\dot{\boldsymbol{r}}_i(t^{\prime\prime})
\end{equation*}
and define
\begin{equation*}
\kappa_i(t^{\prime\prime})
\doteq
1-\frac{1}{c}\boldsymbol{n}_i(t^{\prime\prime})\cdot\dot{\boldsymbol{r}}_i(t^{\prime\prime})
\end{equation*}
such that
\begin{gather*}
\phi_i(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\int\limits_{t^{\prime\prime\prime}}
\frac{q_i\delta\left(t^{\prime\prime\prime}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|\kappa_i(t^{\prime\prime})}
\text{d}{t^{\prime\prime\prime}}
\\
\boldsymbol{A}_i(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\int\limits_{t^{\prime\prime\prime}}
\frac{q_i\dot{\boldsymbol{r}}_i(t^{\prime\prime})\delta\left(t^{\prime\prime\prime}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|\kappa_i(t^{\prime\prime})}
\text{d}{t^{\prime\prime\prime}}
\end{gather*}
and finally the last definition
\begin{gather*}
\tau
+\frac{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|}{c}
\doteq
t
\\
\phi_i(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\frac{q_i}{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|\kappa_i(\tau)}
\\
\boldsymbol{A}_i(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\frac{q_i\dot{\boldsymbol{r}}_i(\tau)}{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|\kappa_i(\tau)}
\end{gather*}
That are the potential of Liénard-Wiechart. The expression of the electric field that you cited is just a consequence of Maxwell equations and so electromagnetic field can be obtained by the potentials, being careful with the gradient and the time derivative. Hope this helps
