$ \frac{\partial \mathscr{L}}{\partial \Phi^i} - \partial_\mu \left(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\Phi^i)}\right) = 0 $
see also Lagrangian Formalism
The basic idea behind the Lagrangian formalism is that nature is guided by a principle of "minimal action". The Euler-Lagrange equations give the path with a minimal amount of "action" that a system follows.
In principle, there are many possible paths how some given particle or multiple particles could get from some point $A$ to another point $B$. The Euler-Lagrange equations are used to calculate the correct path that a particle follows between $A$ and $B$.
The Euler-Lagrange equation tells us which path is the path with minimal action $S = \int_{t_i}^{t_f} dt L(q,\dot{q})$, where $L(q,\dot{q})$ denotes the Lagrangian.
$$ \text{For particles: } \frac{\partial L}{\partial q_i} - \frac{d }{d t}\frac{\partial L}{\partial \dot{q_i}} = 0 . $$
The Euler-Lagrange equation can also be used in a field theory and there it tells us which sequence of field configurations has minimal action.
$$ \text{For fields: } \frac{\partial \mathscr{L}}{\partial \Phi^i} - \partial_\mu \left(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\Phi^i)}\right) = 0 .$$
The general procedure is that we start with a Lagrangian. The Lagrangian is an object that has to be guessed by making use of symmetry considerations and characterizes the system in question. Then we put the Lagrangian into the Euler-Lagrange equation and this gives us the equations of motion of the system.
Derivation
We consider an arbitrary path $(q,\dot{q})$. If it is the path that minimizes the action, we have \begin{eqnarray} 0 &=& \delta S = \delta \int_{t_i}^{t_f} dt L(q,\dot{q}) = \int_{t_i}^{t_f} dt L(q+\delta q,\dot{q}+\delta \dot{q})-S \\ &=& \int_{t_i}^{t_f} dtL(q,\dot{q}) + \int_{t_i}^{t_f} dt\bigg(\delta q {\partial L \over \partial q} + \delta \dot{q} {\partial L \over \partial \dot{q}} \bigg) - S \\ &=& \int_{t_i}^{t_f} dt \bigg(\delta q {\partial L \over \partial q} + {\partial L \over \partial \dot{q}} {d \over dt} \delta q\bigg) \end{eqnarray} If we now integrate the second term by parts, and take the variation of $\delta q$ to be $0$ at $t_i$ and $t_f$, \begin{eqnarray} \delta S = \int_{t_i}^{t_f} dt \bigg(\delta q{\partial L \over \partial q} - \delta q {d \over dt} {\partial L \over \partial \dot{q}} \bigg) = \int_{t_i}^{t_f} dt \delta q \bigg({\partial L \over \partial q} - {d \over dt} {\partial L \over \partial \dot{q}} \bigg) = 0 \end{eqnarray} Now, the only way this holds for an arbitrary variation $\delta q$ of the path $(q,\dot{q})$ is when \begin{eqnarray} {d \over dt}{\partial L \over \partial \dot{q}} - {\partial L \over \partial q} = 0 \end{eqnarray} This is the Euler-Lagrange equation.
Generalized to multiple coordinates $q_i$ ($i=1,\ldots,n$) it reads \begin{eqnarray} {d \over dt} {\partial L \over \partial \d