Hint:
The tangent vector at $\bigl(x(t_0),y(t_0),z(t_0)\bigr)$ is normal to the plane $24x-6y+64z=0$.
So you have to solve for $\;\dfrac{3t_0^2}{24}=\dfrac{3}{6}=\dfrac{4t_0^3}{-64}$.
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?
Best Answer
You are correct that you will need to find $r'(t)$, the derivative of your function. This will give you the direction vector of your curve at an arbitrary time $t$.
From there, all we have to do is determine when this direction vector is perpendicular to the normal of the plane and solve for $t$.
Start by finding $r'(t)$. Simply differentiate each component of $r(t)$ and we get:
$$ r'(t) = <0,1,2t> $$
We can then extract the normal from the plane rather easily, as the normal of a plane in the form $ax + by + cz = 0$ is simply $(a,b,c)$.
Thus our normal vector for our plane is $(1,2,3)$.
To check where this is perpendicular to the direction vector of our curve, we will see where their dot product is $0$:
$$ (0,1,2t)\cdot(1,2,3) = 0 \Rightarrow 0 + 2 + 6t = 0 \Rightarrow 6t = -2 \Rightarrow t = -2/6 = -1/3 $$
Now all we have to do is plug in $t = -1/3$ to the equation of our curve to find the given point: $r(-1/3) = (1, -1/3, 1/9)$.