What is the condition of a curve to have a tangent at point P? 
A curve in $R^n$ is a continuous map $\gamma :[a,b] \to R^n$.
The image of $\gamma$ is often called curved,too. When there is need to distinguish between different curves with the same image, different functions $\gamma$ are referred to as parameterizations.

For this part, I don't understand why different functions are referred to as parameterizations? Can anyone give me some examples?

A curve (for simplicity, in $R^3$) passing through a point P is said to have a tangent at P if there is a parametrization$(\xi(t),\eta(t),\zeta(t))$ such that $(\xi^{'}(t))^2 +(\eta^{'}(t))^2+(\zeta^{'}(t))^2$ $\ne 0$
at value of parameter t corresponding to P. In this event, vector $(\xi^{'}(t),\eta^{'}(t),\zeta^{'}(t))$ is called a tangent vector to $\gamma$ at P.

For this part , I don't know why

$(\xi^{'}(t))^2 +(\eta^{'}(t))^2+(\zeta^{'}(t))^2$ $\ne 0$ $\qquad$  (1)

is necessary for a curve to have a tangent at P.  In $R^2$, doesn't it is enough for a  curve have a tangent if the curve is differentiable at P? Can anyone give me some examples about this condition?
 A: 
Why are different functions referred to as parametrizations?

If you've learned about parametric equations, you know that we can define a curve in the plane using a pair of equations $x(t),\ y(t)$. The variable $t$ is not associated with any spatial axis (like $x$ or $y$), but instead is, well, a parameter that $x$ and $y$ can vary over so that they trace out a curve. Hence the term parametrization.$\newcommand{\R}{\mathbb R}$
The reason the notion of parametrization comes up in your case is, I think, because different curves (in the sense of functions from $[a,b] \to \R^n$) can have the same images (i.e. graphs in $\R^2$ or $\R^3$ or $\R^n$). As an example, consider this curve:
\begin{align}
x(t) &= t \\
y(t) &= t^2
\end{align}
If you graph it, you'll see it generates the curve $y=x^2$. Now consider this curve:
\begin{align}
x(t) &= 2t \\
y(t) &= 4t^2
\end{align}
This is definitely a different "curve" than the first one because the functions are different; and yet if you plot it, you get the same graph out: namely $y = x^2$. So we have two different "curves" that "look" identical.
When this happens, we distinguish between the two sets of functions by saying that $(t, t^2)$ and $(2t, 4t^2)$ are different "parametrizations" of the same "curve". The thing you have to remember, though, is sometimes the word "curve" is used interchangeably with a particular parametrization of a curve! That can make things confusing, but usually good authors should only interchange the two if there's no chance of a conflict happening between the two senses of the word "curve".

Why the requirement that $\xi^{'}(t)^2 + \eta^{'}(t)^2 + \zeta^{'}(t)^2 \neq 0$?

Basically this is saying that a curve is said to have a tangent at a point $P$ if and only if there exists any parametrization of the curve that has a non-zero derivative at $P$.
To see why this condition is necessary, consider the curve $y = |x|$. You may remember that it looks like a V with the vertex located at the origin. Obviously, this curve has no tangent at the origin, but I could define a parametrization that makes it look like it does. For example, this one:
\begin{align}
x(t) &= t^3 \\
y(t) &= |t^3|
\end{align}
If you graph it, you'll see that this is a valid parametrization of $y = |x|$, but if you take the derivative of both at $t = 0$ you'll get that
\begin{align}
x'(0) &= 0 \\
y'(0) &= 0
\end{align}
(If you don't believe me, try graphing $x$ vs $t$ and $y$ vs $t$ on their own separate axes. You'll see they both go flat at $t=0$.)
The non-zero requirement prevents us from incorrectly concluding this curve has a tangent at the origin.
As a bit of an aside, you can also use these kinds of "bad" parametrizations to make it look like a curve that does in fact have a tangent doesn't. Consider this parametrization of $y=x^2$:
\begin{align}
x(t) &= t^3 \\
y(t) &= t^6
\end{align}
Both of these functions have zero derivative at $t = 0$. Does that mean the curve has no tangent at the origin? No. Because the definition states that if there exists ANY parametrization that has a non-zero derivative at a point, then that point has a tangent. Above, we just happened to find a parametrization that didn't work, but we can easily find one that does:
\begin{align}
x(t) &= t \\
y(t) &= t^2
\end{align}
The thing with $y = |x|$ is that you won't ever be able to find any differentiable parametrization that is non-zero at the origin. Either it won't be differentiable there, or it will be, but it'll be zero.

UPDATE: In response to the following question posted in a comment:

Is it impossible to say that a function doesn't have a tangent at a point by using this condition? Because it is impossible to go through ALL parametrizations.

If I understand you correctly, I think the answer is Yes: You can't use the non-zero derivative condition with only one parametrization to prove a particular curve has no tangent. You really would need to check all possible parametrizations. That sounds like an impossible task, but we can check all possibilities all at once by letting $x(t), y(t)$ represent an arbitrary parametrization.
To see how, here's a sketch of the proof of why $y=|x|$ has no tangent at the origin:
Let $x(t), y(t)$ be any parametrization of $y=|x|$, and suppose that both functions are differentiable at the point $t_0$ which satisfies $x(t_0) = y(t_0) = 0$ (i.e. maps the parametrization to the origin). We want to show that the derivatives of both must be zero (and hence fail to satisfy the definition of having a tangent). To do that we'll consider whether either of them could possibly be non-zero.
If $y'(t_0) \neq 0$, then that means that $y$ is not flat when it is crossing the $t$ axis: it has some slope at a zero. That means $y$ must be negative at some $t$ values near $t_0$. That is not possible since, being a parametrization of $y=|x|$, it must always be positive. This eliminates the possibility that $y'(t_0)$ could be non-zero.
Consider the case $x'(t_0) \neq 0$. Note that since $y(t) = |x(t)|$, we can take the derivative of both sides at the point $t_0$ and conclude:
\begin{align}
y'(t_0) &= \lim_{h \to 0} \frac{|x(t_0+h)| - |x(t_0)|}{h} \\
&= \lim_{h \to 0} \frac{|x(t_0+h)|}{h} \quad \text{since $x(t_0)=0$ by definition}
\end{align}
By the same reasoning made before, since $x'(t_0) \neq 0$, $x$ must be positive near $t_0$ on one side, and negative on the other side. For sake of simplicity, I'll assume that $x$ is positive when $t$ is a little bit larger than $t_0$. Now, since $y'(t_0)$ exists, that means it must equal the left- and right-hand limits. In particular, it must equal the right-hand limit:
\begin{align}
y'(t_0) &= \lim_{h \to 0^+} \frac{|x(t_0 + h)|}{h} \\
&= \lim_{h \to 0^+} \frac{x(t_0+h)}{h} \quad \text{because $x$ is positive to the right of $t_0$} \\
&= x'(t_0) \neq 0
\end{align}
But this is a contradiction because we've already shown that $y(t_0)$ can't be non-zero. Hence $x(t_0)$ can't be non-zero either. $\square$
