Full time-derivative, Poisson brackets and Hamilton's equations (classical mechanics) While studying Poisson brackets in classical mechanics and the derivation of $\dot{q_j}=\{q_j,H\}$ and $\dot{p_j}=\{p_j,H\}$ form of Hamilton's equations I encountered a surpsing identity, which led me to think that maybe I got something wrong about full time-derivative, which is as follows:
$$
\frac{d\,f(q_1,q_2,\dots,q_N,p_1,p_2,\dots,p_N,t)}{dt}=\sum_{j=1}^{N}\left(
\frac{\partial f}{\partial q_j}
\underbrace{\frac{\partial q_j}{\partial t}}_{=\dot{q_j}=\frac{\partial H}{\partial p_j}}+
\frac{\partial f}{\partial p_j}
\underbrace{\frac{\partial p_j}{\partial t}}_{=\dot{p_j}=-\frac{\partial H}{\partial q_j}}
\right)+\frac{\partial f}{\partial t}
=\{f,H\}+\frac{\partial f}{\partial t}
$$
Now if for example I use function $f$ as $q_k$, by saying: $f(q_1,q_2,\dots,q_N,p_1,p_2,\dots,p_N,t)=q_k$ then I get following:
$$
\frac{dq_k}{dt}=\sum_{j=1}^{N}\left(
\underbrace{\frac{\partial q_k}{\partial q_j}}_{=\delta_{kj}}
\frac{\partial H}{\partial p_j}-
\underbrace{\frac{\partial q_k}{\partial p_j}}_{=0}
\frac{\partial H}{\partial q_j}
\right)+\frac{\partial q_k}{\partial t}
=\{q_k,H\}+\frac{\partial q_k}{\partial t}
=
\underbrace{\frac{\partial H}{\partial p_k}}_{=\dot{q_k}}
+
\underbrace{\frac{\partial q_k}{\partial t}}_{\stackrel{?}{=}\dot{q_k}}\stackrel{?}{=}2\dot{q_k}
$$
If I cross out the $\frac{\partial q_k}{\partial t}$ on both sides of equation $\{q_k,H\}+\frac{\partial q_k}{\partial t}=\frac{\partial H}{\partial p_k}+\frac{\partial q_k}{\partial t}$ then I recover the Hamilton's equation $\dot{q_k}=\{q_k,H\}$. But if I don't do this, and go forward with the $2\dot{q_k}$ that appears at the end I get this very surprising identity:
$$
\frac{dq_k}{dt}\stackrel{?}{=}2\dot{q_k}
$$
or written another way:
$$
\frac{dq_k}{dt}\stackrel{?}{=}2\frac{\partial q_k}{\partial t}
$$
My question is following: is this really true? If not, then what have I done wrong? If yes, then why it is not mentioned anywhere in the textbooks - wouldn't that be some other way to find the $\dot{q_k}$?
Note:The derivation goes exactly the same for $\dot{p_k}$.

EDIT: Thanks to your answers I marked now the wrong equalities with $\stackrel{?}{=}$ since I hate to see incorrect math typed out. But I still wanted to preserve this question as how it was written at first. Also in the first equation the full time derivative for $\dot{q_j}$ and $\dot{p_j}$ should be used like this:
$$
\frac{df}{dt}=\sum_{j=1}^{N}\left(
\frac{\partial f}{\partial q_j}
\underbrace{\frac{d q_j}{d t}}_{=\dot{q_j}=\frac{\partial H}{\partial p_j}}+
\frac{\partial f}{\partial p_j}
\underbrace{\frac{d p_j}{d t}}_{=\dot{p_j}=-\frac{\partial H}{\partial q_j}}
\right)+\frac{\partial f}{\partial t}
$$
 A: The problem is that you are equating too many things to $\dot{q_k}$.
Usually $\dot{q_k} = \frac{dq_k}{dt}$, a total derivative, as opposed to a partial derivative.
If $q_k$ has no explicit time-dependence, i.e. it does not depend directly on $t$ itself, then $\frac{\partial q_k}{\partial t} = 0.$
In this case, the Poisson bracket reduces to: $ \frac{dq_k}{dt} = \frac{\partial H}{\partial p_k} $, so now you can say $\dot{q_k} = \frac{\partial H}{\partial p_k} $.
If you now consider your $q_k$ to have a time dependence, so $q_k(t)$, the Poisson bracket becomes, as you have pointed out:  $ \frac{dq_k}{dt} = \frac{\partial H}{\partial p_k}+\frac{\partial q_k}{\partial t}$, so $ \dot{q_k} =\frac{\partial H}{\partial p_k} +\frac{\partial q_k}{\partial t} $.

ADDITION after discussion in comments:
$\frac{d}{dt}f$ is a total derivative with respect to time, which means it picks out ALL time dependences of $f(t, x(t), y(t), z(t))$. Collectively, call $x,y,z$ as $\{x_i\}$.
Using the chain rule, $$ \frac{d}{dt} = \frac{\partial}{\partial x_i}\frac{\partial x_i}{\partial t} = \frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial}{\partial z}\frac{\partial z}{\partial t}+\frac{\partial}{\partial t}\underbrace{\frac{\partial t}{\partial t}}_1 = \mathbf{u}\cdot\mathbf{\nabla}+\frac{\partial}{\partial t}.$$
In Hamiltonian mechanics, you parameterise a function in terms of its positions $\{q_i\}$ and its momenta $\{p_i\}$. Basically the $\{x_i\}$ from earlier are now $\{q_i, p_i\}$.
$$ \frac{df}{dt}=\frac{\partial f}{\partial x_i}\frac{\partial x_i}{\partial t} = \frac{\partial f}{\partial q_i}\frac{\partial q_i}{\partial t}+\frac{\partial f}{\partial p_i}\frac{\partial p_i}{\partial t} + \frac{\partial f}{\partial t}.$$
The definition of the Poisson bracket is:
$$ \{f,H\}=\left(\frac{\partial f}{\partial q_i}\frac{\partial H}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial H}{\partial q_i} \right). $$
Plugging in the Hamilton's equations:
$$ \{f,H\}=\left(\frac{\partial f}{\partial q_i}\frac{dq_i}{dt}+\frac{\partial f}{\partial p_i}\frac{dp_i}{dt} \right), $$ when $q_i$ and $p_i$ are the fundamental variables, so they are not functions of $x, y, z$ but only time. This means that, applying the formula from before, $\frac{\partial q_i}{\partial t} = \frac{dq_i}{dt}$, so:
$$ \{f,H\}=\left(\frac{\partial f}{\partial q_i}\frac{\partial q_i}{\partial t}+\frac{\partial f}{\partial p_i}\frac{\partial p_i}{\partial t} \right). $$
As you can see $\frac{\partial f}{\partial t}$ does not enter the definition.
In conclusion, 
$$ \frac{df}{dt}=\{f,H\}+\frac{\partial f}{\partial t}. $$
P.S. I have been using  Einstein's summation convention: repeated indeces imply summation. $$\frac{\partial f}{\partial p_i}\frac{\partial p_i}{\partial t} = \sum_{i=1}^{N} \frac{\partial f}{\partial p_i}\frac{\partial p_i}{\partial t}.$$
