My professor wrote this $\frac{\partial L}{\partial q}\dot{q}=\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})$. Due to the fact that I am very very very very bad at Math, could you explain me about this please? I don't understand.
$L(q,\dot{q},t)$ is the Lagrangian.
First, the problem you are trying to solve is:
$$ S = \int\limits_{t_1}^{t_2} \mathcal{L}(q, \dot{q}, t)dt $$
Assume there is a particular function, $q_0(t)$, which minimizes* this integral. We perturb this function and find the value of the integral with the perturbed functions:
$$ q(\eta, t) = q_0(t) + \eta \lambda(t) $$
All that we require is that $\lambda(t_1) = \lambda(t_2) = 0$ (this function vanishes at the end points of the integral). This gives, for any particular function $\lambda(t)$ that satisfies our boundary conditions:
$$ S(\eta) = \int\limits_{t_1}^{t_2} \mathcal{L}(q_0 + \eta\lambda, \dot{q}_0 + \eta\dot{\lambda}, t)dt $$
The tricky part (for me) is understanding that you can push the differentiation of $S$ inside of the integral:
$$ \frac{dS}{d\eta} = \int\limits_{t_1}^{t_2} \frac{d\mathcal{L}(q_0 + \eta\lambda, \dot{q}_0 + \eta\dot{\lambda}, t)}{d\eta}dt = 0 \\ \frac{dq}{d\eta} = \lambda, \frac{d\dot{q}}{d\eta} = \dot{\lambda}, \frac{dt}{d\eta} = 0 $$
This can also be confusing: why are are we trying to solve $\frac{dS}{d\eta} = 0$? The answer is that we're trying to find a critical point for the function $S(\eta)$, given some function $\lambda(t)$ (that satisfies our boundary values). What we find below is that $\eta$ drops out of the equation and thus also our extraneous $\lambda$ function. What we eventually find (assuming it's a minimum), is give some random $\lambda(t)$, this function $q(t)$ will give the "minimum" given the random $\lambda(t)$ function. We can find the "absolute" minimum by simply setting $\lambda(t) = 0$. We don't need to prove that $\eta = 0$ means a critical point--for one, we don't know what $q_0$ is, and second, the idea is that given any $\eta$ and $\lambda(t)$, this optimal $q(t)$ will give the "lowest" value!
$\eta$ drops out specifically because we've chosen $\eta$ to be a linear dependence--had we chosen $\eta^2$, this doesn't work. So then you ask, is this general? Well yes because we've allowed any $\lambda(t)$ function so long as it satisfies the boundary conditions ($0$ at both ends)--meaning it can do anything, even mimic $\eta^2$ between the boundaries!
\begin{align*} \frac{d\mathcal{L}}{d\eta} =&\ \frac{\partial \mathcal{L}}{\partial q}\frac{dq}{d\eta} + \frac{\partial \mathcal{L}}{\partial \dot{q}}\frac{d\dot{q}}{d\eta} + \frac{\partial\mathcal{L}}{\partial t}\frac{dt}{d\eta}\\ =&\ \frac{\partial \mathcal{L}}{\partial q}\lambda + \frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda} = 0 \end{align*}
$$ \int\limits_{t_1}^{t_2}\left(\frac{\partial \mathcal{L}}{\partial q}\lambda + \frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda}\right)dt = 0 \\ \int\limits_{t_1}^{t_2}\frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda}dt = -\int\limits_{t_1}^{t_2}\frac{\partial \mathcal{L}}{\partial q}\lambda dt $$
Do integration by parts (i.e. the product rule) for the left integral:
\begin{align} u =& \frac{\partial \mathcal{L}}{\partial \dot{q}} & du =& \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right)dt \\ dv =& \dot{\lambda}dt & v =& \lambda \end{align}
$$ \int\limits_{t_1}^{t_2} \frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda}dt = \left.\lambda \frac{\partial \mathcal{L}}{\partial \dot{q}}\right|_{t_1}^{t_2} - \int\limits_{t_1}^{t_2} \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) \lambda dt $$
But we said $\lambda(t_1) = \lambda(t_2) = 0$ so the first part of this integral vanishes, leaving:
$$ \int\limits_{t_1}^{t_2} \frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda}dt = - \int\limits_{t_1}^{t_2} \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) \lambda dt $$
Substitute this back into the equation $\int\frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda}dt = -\int\limits\frac{\partial \mathcal{L}}{\partial q}\lambda dt$
$$ - \int\limits_{t_1}^{t_2} \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) \lambda dt = -\int\limits_{t_1}^{t_2}\frac{\partial \mathcal{L}}{\partial q}\lambda dt \\ \int\limits_{t_1}^{t_2} \left(\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) - \frac{\partial \mathcal{L}}{\partial q}\right)\lambda dt = 0 $$
For this integral to be zero for all functions $\lambda$, the thing multiplying $\lambda$ must be zero, hence the Euler-Lagrange equation:
$$ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) - \frac{\partial \mathcal{L}}{\partial q} = 0 $$
*I say "minimize"--this is not correct. It's not even correct to say "find extremum". In reality, we're simply finding a function that is analogous to a critical point. And likewise, once we find such a function, we must verify it's the extremum we're looking for.