Finding the Tangent Line to a Surface 
What is the equation of the tangent line to the intersection of the surface $z = \arctan (xy)$ with the plane $x=2$, at the point $(2,\frac{1}{2}, \frac{\pi}{4})$

The intersection of $x=2$ and $z= \arctan (xy)$ produces the curve $z = \arctan (2y)$ in the $yz$-plane. Thus, the partial derivative is $\frac{\partial z }{\partial y} = \frac{\partial z }{\partial y} \arctan(2y) = \frac{2}{1 +4y^2}$ and the slope of the line passing through $y=1/2$ is $\frac{\partial z }{\partial y} = 1$
My first question is, is there a conceptual/logical error in plugging the $x=2$ before taking the (partial) derivative. In the book I am using for review, the author computes the partial derivative of $z= \arctan (xy)$ wrt $y$ and then plugs in the point $(2, \frac{1}{2})$. I realize that I obtained the same answer, but that does not necessarily imply it's a valid way of solving the problem.  
Now, $\frac{\partial z }{\partial y} = 1$ gives us the slope of the 2-D version of the line whose equation we interested in finding. Having a little trouble determining the equation, I consulted the book and this is what it says: 

"Since tangent line is in the plane $x=2$, this calculation [namely, the calculation of $\frac{\partial z }{\partial y}$] shows that the line is parallel to the vector $v = (0,1,1)$." 

I don't see how it follows from $\frac{\partial z}{\partial y} = 1$ that the "slope" vector (not exactly sure what it is called) is $v=(0,1,1)$. 
 A: For your first question:

is there a conceptual/logical error in plugging the $x=2$ before taking the (partial) derivative.

No, and the reason is that you're plugging in a value for $x$ and you're taking the partial derivative with respect to (wrt) $y$.  When you take the partial wrt $y$, you treat $x$ as a constant anyway.  So it won't matter if you replace $x$ with a constant before or after taking the partial wrt $y$.  Note that it will matter if you take the partial wrt $x$ instead, because replacing $x$ with a constant will force the partial wrt x to be zero.
For your second question:

I don't see how it follows from $\frac{\partial z}{\partial y} = 1$ that the "slope" vector (not exactly sure what it is called) is $v=(0,1,1)$. 

The vector $v$ doesn't have any special name with respect to $\frac{\partial z}{\partial y}$.  It's simply a vector that's parallel to the tangent line.  Anyway, the calculation gives us
$$
  \frac{\partial z}{\partial y} = \frac2{4y^2+1}.
$$
And remember we're dealing with the tangent line at the point $(2, 1/2, \pi/4)$.  So $y = 1/2$, which means
$$
  \frac{\partial z}{\partial y}\bigg|_{(x,y,z) = (2,1/2,\pi/4)} = \frac2{4(\frac12)^2+1} = 1.
$$
So the line has slope $1$ and passes through $(x,y,z) = (2,\frac12, \frac\pi4)$.  Since we can view this as a 2D line because we're working in the plane $x=2$, then the equation of our line is $z - \frac\pi4 = 1(y - \frac12)$, i.e., $$z = y + \frac\pi4 - \frac12.$$
So, why is this parallel to the vector $v = \langle 0,1,1 \rangle$?  Notice that the line has slope $1$.  That is, every time $y$ increases by $1$ unit, $z$ also increases by $1$ unit.  The same is true of the vector $v = \langle 0,1,1 \rangle$.  Recall that vectors really only tell us two things:  magnitude and direction.  Location is irrelevant, so for simplicity we can assume vectors begin at the origin.  Then we can view $v = \langle 0,1,1 \rangle$ as the line segment between $(0,0,0)$ and $(0,1,1)$.  (Technically it's a directed line segment but that's not relevant here.)  So we can actually view our vector $v$ as a piece of the line $z=y$ in the plane $x=0$.  Do you agree that $z=y$ and $z=y+ \frac\pi4 - \frac12$ are parallel?
