From the Princeton book for the GRE Subject Test in Maths:
I want to understand how the Princeton book came up with $[0 \ 1 \ 1]$ before the parameterisation. This is how I answered it:
On the plane $x=2$, the slope of the tangent line to $z=\arctan(2y)$ at $(\frac12, \frac\pi4)$ is 1.
The equation of the tangent line (which lies on the plane $x=2$) is $z=my+b$ where $m=1$ at $(y,z) = (\frac12, \frac\pi4)$. Hence, it is $z=1y+\frac\pi4 - \frac12$.
In xyz-space, the equation of the tangent line is
$$x=2$$ $$z=y+\frac\pi4 - \frac12$$
or parametrically
$$x=2$$ $$y=s$$ $$z=s+\frac\pi4 - \frac12$$
where $s=\frac12 + t$
or
$$[x \ y \ z] = \color{red}{[0 \ 1 \ 1]}s + [2 \ 0 \ \frac\pi4 - \frac12]$$
The parametric equation of the curve is $<2,t,arctan(2t)>$ Its derivative is thus $<0,1,\frac{2}{1+4t^2}>$. At the given point (for t=1/2) the direction is $<0,1,1>$. From here we get $x=2,y=1/2+t,z=\pi/2+t$ which is the parametric equation of the line. I think this way is easier than the approach of the book.