The question is the following:
What's the equation of the line tangent 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 explanation in the book is
Since $x$ is held constant, the slope of the tangent line we're looking for is equal to the value of the partial derivative $\frac{\partial}{\partial y}\arctan(xy)$ at $(x_o,y_o)$ = $(2, \frac{1}{2})$:
$\frac{\partial}{\partial y}(\arctan (xy)) = \frac{x}{1 + (xy)^2} \rightarrow slope = [\frac{x}{1 + (xy)^2}]_{(2,\frac{1}{2})} = 1$
Since the tangent line is in the plane $x=2$, this calculation shows that the line is parallel to the vector $v = (0,1,1)$.
What I don't understand is how one gets the vector $v$?
Vectort $\vec v $ could be any multiple of (0,1,1). The 0 is because you are constrained in the $x=2$ plane, so the x value doesn't change. The last two have to match because the slope is one and for every step forward you need to take one step up.
Hope this helps.
Ced