Taking a derivative from both sides of the equation Our teacher tried to explain to us how to find a slope at a given point of the function $y^2 = 2px$ by taking derivative from both sides of the equation, he did this:
$$y^2= 2px $$
$$2yy' = 2p$$
$$ y' = \frac{p}{y}$$
I haven't quite understood the second part of the process, will be thankful for a logical and clear explanation :)
 A: When we find derivative of some term. First we find the derivative of the power then the derivative of term.  
So left hand side of second step,
$\frac{d}{dx}y^2 = 2y^{2-1}.\frac{d}{dx}y$
= $2y^{1}.\frac{dy}{dx}$
And you can write $\frac{dy}{dx} = y'$
So we have,
= $2y.y'$
In second step on the right hand side,
$\frac{d}{dx}(2px)$ 
= 2p$\frac{d}{dx}x$ 
As 2 and p is constant and derivative of x with respect to x is 1.
= 2p
At last,
2y.y' = 2p.
$y' = \frac{p}{y}$
A: It will become clear if you know the concept of implicit differentiation.   
Thus we have that $$y^2=2px $$ $$\Rightarrow 2y\frac {dy}{dx} =2p\frac {dx}{dx} $$ $$\Rightarrow 2yy'= 2p $$ $$\Rightarrow y'=\frac {p}{y} $$ where $y^{(n)} $ is the nth derivative of $y$ (we generally use $y', y''$ etc. for lower derivatives but it is not advisable for higher ones). Hope it helps. 
A: This is just the chain rule you presumably learned about earlier in calc, but the notation is more compact. Recall the chain rule says that
$$
\frac{d}{dx}f(y(x)) = f'(y(x))y'(x)
$$
The LHS of the second line is just the chain rule applied for $f(y) = y^2$. Since $f'(y) = 2y,$ we have 
$$
\frac{d}{dx}(y(x))^2 = 2y(x)y'(x).
$$
A: $y$ is a function of $x$. $y^2$ is differentiated with respect to independent variable $x$ using Chain Rule. Instead of everytime writing like the following
$$y(x)^2= 2px, \,2y(x)y'(x) = 2p,\, y^{\prime}(x) = \frac{p}{y(x)}$$ 
your teacher has omitted repetitive parentheses $(x)$, assuming that you can appreciate that $p$ is a constant and $x,y$ are variables.
BTW $p$ is double the focal length of parabola which has a property that sub-normal ( projection on x-axis $=p)$ is always constant for the normal (in red).

