Proper time along path in Minkowski Space Consider the path $x^\mu(u)$ in Minkowski space; such that:
$$t = \frac{a}{c} \sinh(u) , \quad x = a \cosh(u) ,\quad y = 0 ,\quad z = 0 $$
where $a$ is a positive constant and $u$ is a parameter
Use equation: 
$$ c \nabla \tau = ds = \sqrt{\eta_{\mu\nu} \dot{x}^\mu \dot{x}^\nu} du  $$
 to find the proper time elapsed along the path starting from $u = 0$, as
a function of $u$.
I'm having quite a bit of trouble with using the notation if someone could help me please?
So far I have established:
$$ d_u x^{\mu} = \left(\frac {a}{c}\cosh(u),\, a\sinh(u),\,0,\,0\right)) $$
Then:
$$\eta_{\mu\nu} d_u x^{\mu} d_u x^{\nu} = \sum_{\mu =0}^{3} \sum_{\nu =0}^3 \eta_{\mu\nu} d_u x^{\mu} d_u x^{\nu}  $$
Then I get a bit lost...
 A: Let us assume that you use the metric:
$$\left\{ \eta_{\mu \nu} \right\} =  \begin{pmatrix} 1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\ 
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1 \end{pmatrix}
$$
(Note that you could also use:
$$\left\{ \eta_{\mu \nu} \right\} =  \begin{pmatrix} -1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{pmatrix}
$$
which is just a matter of conventions.)
Now, we interpret the notations as follows (as you have mentioned in your comment):
\begin{equation}
\sqrt{\eta_{\mu \nu} \dot{x}^\mu\dot{x}^\nu} =  \sqrt{\sum\limits_{\mu=0}^3\sum\limits_{\nu=0}^3\eta_{\mu \nu} \dot{x}^\mu \dot{x}^\nu} \tag{1}
\end{equation}
In order to evaluate this, it is important to realize:
$$
 \eta_{\mu \nu} = 0 \text{ if } \mu \neq \nu$$
Therefore, writing out equation $(1)$ fully:
\begin{equation}
\begin{aligned}
\sqrt{\eta_{\mu \nu} \dot{x}^\mu \dot{x}^\nu} & = \sqrt{ \eta_{00}\dot{x}^0 \dot{x}^0 + \eta_{11}\dot{x}^1 \dot{x}^1 + \eta_{22}\dot{x}^2 \dot{x}^2 +\eta_{33}\dot{x}^3 \dot{x}^3} \\&
= \sqrt{ \dot{x}^0 \dot{x}^0 - \dot{x}^1 \dot{x}^1 -\dot{x}^2 \dot{x}^2 -\dot{x}^3 \dot{x}^3}
\end{aligned}
\end{equation}
Furthermore:
\begin{equation}
\left\{ x^\mu \right\} = \begin{pmatrix} x^0 \\
x^1 \\ 
x^2 \\
x^3 \end{pmatrix} = \begin{pmatrix} ct \\
x \\ 
y \\
z \end{pmatrix} \tag{2}
\end{equation}
At this point it is just substituting everything in the formula and evaluating it.
Edit:
One of the most important properties of special relativity is that all inertial reference frames are physically equivalent. This means that if one observer sees $ct,x,y,z$ and the other sees $ct',x',y',z'$, which may in general not be equal, we still have:
\begin{equation}
c^2 t^2-x^2 - y^2 - z^2 = c^2 t'^2-x'^2 - y'^2 - z'^2
\end{equation}
This can be written more compactly (and manifestly Lorentz invariant) by the way equation $(2)$ is defined:
$$ \eta_{\mu \nu} x^\mu x^\nu = \eta_{\mu \nu} x'^\mu x'^\nu $$
This is why equation $(2)$ is defined the way it is.
