Here's a nice little Mathematica routine for evaluating Tito's continued fraction with precision prec:
prec = 10^4;
y = N[4, prec];
c = y; d = 0; k = 1;
u = 1; v = y;
While[True,
c = 1 + u/c; d = 1/(1 + u d);
h = c*d; y *= h;
v += 96 k^2 + 8;
c = v + u/c; d = 1/(v + u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-prec, Break[]];
u += 3 k (k + 1) + 1;
k++];
6/y
where I used the Lentz-Thompson-Barnett method for the evaluation.
For prec = 10^4, the thing evaluates in 120 seconds (via AbsoluteTiming[]), giving a result that agrees with $\zeta(3)$ to 10,000 digits.
One can consider the even part of Tito's CF, which converges at twice the rate of the original:
$$\cfrac{6}{5-\cfrac{u_1}{v_1-\cfrac{u_2}{v_2-\cfrac{u_3}{v_3-\cdots}}}}$$
where
$$\begin{align*}
u_k&=k^6\\
v_k&=(17k^2+17k+5)(2k+1)
\end{align*}$$
Here's Mathematica code corresponding to this CF:
prec = 10^4;
y = N[5, prec];
c = y; d = 0; k = 1;
While[True,
u = k^6;
v = (2 k + 1) ((17 k + 17) k + 5);
c = v - u/c; d = 1/(v - u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-prec, Break[]];
k++];
6/y
For prec = 10^4, the thing evaluates in 70 seconds (via AbsoluteTiming[]). There may be further ways to accelerate the convergence of the CF, but I have yet to look into them.
Added, quite a bit later:
As it turns out, the even part I derived is precisely Apéry's CF for $\zeta(3)$ (thanks Américo!). Conversely put, Tito's CF is an extension of Apéry's CF. Here's how to derive Apéry's CF from Tito's CF (while proving convergence along the way).
We start from an equivalence transformation of Tito's CF. A general equivalence transformation of a CF
$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cfrac{a_3}{b_3+\cdots}}}$$
with some sequence $\mu_k, k>0$ looks like this:
$$b_0+\cfrac{\mu_1 a_1}{\mu_1 b_1+\cfrac{\mu_1 \mu_2 a_2}{\mu_2 b_2+\cfrac{\mu_2 \mu_3 a_3}{\mu_3 b_3+\cdots}}}$$
Now, given a CF
$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cdots}}$$
one can transform this into a CF of the form
$$b_0+\cfrac{w_1}{1+\cfrac{w_2}{1+\cdots}}$$
where $w_1=\dfrac{a_1}{b_1}$ and $w_k=\dfrac{a_k}{b_k b_{k-1}}$ for $k > 1$, where we used $\mu_k=\dfrac1{b_k}$. Applying this transformation to Tito's CF yields the CF
$$\cfrac{\frac32}{1+\cfrac{w_2}{1+\cfrac{w_3}{1+\cdots}}}$$
where $w_{2k}=\dfrac{k^3}{4(2k-1)^3}$ and $w_{2k+1}=\dfrac{k^3}{4(2k+1)^3}$. (You can easily demonstrate that this transformed CF and Tito's CF have identical convergents.)
At this point, we find that since the $w_k \leq\dfrac14$, we have convergence of the CF by Worpitzky's theorem.
Now, we move on to extracting the even part of this transformed CF. Recall that if a CF has the sequence of convergents
$$u_0=b_0,u_1=b_0+\cfrac{a_1}{b_1},u_2=b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2}},\dots$$
then the even part is the CF whose convergents are $u_0,u_2,u_4,\dots$ (Analogously, there is the odd part with the sequence of convergents $u_1,u_3,u_5,\dots$)
Now, given a CF of the form
$$b_0+\cfrac{w_1}{1+\cfrac{w_2}{1+\cdots}}$$
its even part is the CF
$$b_0+\cfrac{w_1}{1+w_2-\cfrac{w_2 w_3}{1+w_3+w_4-\cfrac{w_4 w_5}{1+w_5+w_6-\cdots}}}$$
Thus, the even part of the previously transformed CF is given by
$$\cfrac{\frac32}{\frac54-\cfrac{\beta_1}{\delta_1-\cfrac{\beta_2}{\delta_2-\cdots}}}$$
where
$$\begin{align*}
\beta_k&=\frac{k^3}{4(2k-1)^3}\frac{k^3}{4(2k+1)^3}=\frac{k^6}{16(2k-1)^3(2k+1)^3}\\
\delta_k&=1+\frac{k^3}{4(2k+1)^3}+\frac{(k+1)^3}{4(2k+1)^3}=\frac{17k^2+17k+5}{4(2k+1)^2}
\end{align*}$$
We're almost there! We only need to perform another equivalence transformation, which I'll split into two steps to ease understanding. First, the easy one with $\mu_k=4$, which yields the CF
$$\cfrac{6}{5-\cfrac{16\beta_1}{4\delta_1-\cfrac{16\beta_2}{4\delta_2-\cdots}}}$$
The last step is to cancel out the odd integer denominators of the $\beta_k$ and $\delta_k$; to do this, we take $\mu_k=(2k+1)^3$; this finally yields the CF
$$\cfrac{6}{5-\cfrac{u_1}{v_1-\cfrac{u_2}{v_2-\cfrac{u_3}{v_3-\cdots}}}}$$
where
$$\begin{align*}
u_k&=k^6\\
v_k&=(17k^2+17k+5)(2k+1)
\end{align*}$$
and this is Apéry's CF.
For completeness, I present a formula for the odd part of Tito's CF, after some post-processing with a few equivalence transformations:
$$\zeta(3)=\frac32-\cfrac{81}{\lambda_1-\cfrac{\eta_1}{\lambda_2-\cfrac{\eta_2}{\lambda_3-\ddots}}}$$
where
$$\begin{align*}
\eta_k&=4\times(4k^4+8k^3+k^2-3k)^3=4\times10^3,\,4\times126^3,\dots\\
\lambda_k&=8\times(68k^6-45k^4+12k^2-1)=8\times34,\,8\times3679,\dots
\end{align*}$$
The formula is somewhat more complicated, and converges at the same rate as the even part.
I found this interesting research that I share with you in case you are interested in the relationship between the negative and positive continued fractions:
https://scholarship.claremont.edu/cgi/viewcontent.cgi?filename=2&article=1183&context=hmc_theses&type=additional
According to Definition 8, the first terms of the negative continued fraction of the Euler Mascheroni constant would be:
$$\gamma=[1,3,2,3,2,3,2,2,2,5,2,2,2,2,2,2,2,2,2,2,2,2,7,3,2,2,2,2,2,2,2,3,2,...]$$
Best Answer
The continued fraction representation above had its origins on another problem I was working on sometime ago.
It's based on a very simple way of looking at the Euler's product representation of $\frac{1}{\zeta(s)}$. Interestingly it applies to every infinite product.
And this is as follows
$$ \frac{1}{\zeta(s)}=\left(1-\frac{1}{2^s}\right)-\left(1-\frac{1}{2^s}\right)\frac{1}{3^s}-\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\frac{1}{5^s}-\cdots $$
From here its easy to derive the above continued fraction using Euler's continued fraction formula.
And thats it, It's nice and eventually a new thing.
EDIT
Just to make it clear, note that $$ \begin{align*} \frac{1}{\zeta(s)}&=\left(1-\frac{1}{2^s}\right)\left[\left(1-\frac{1}{3^s}\right)-\left(1-\frac{1}{3^s}\right)\frac{1}{5^s}-\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{5^s}\right)\frac{1}{7^s}-\cdots\right]\\ &=\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\left[\left(1-\frac{1}{5^s}\right)-\left(1-\frac{1}{5^s}\right)\frac{1}{7^s}-\cdots\right]\\ &\vdots\\ &=\prod_{p\in\mathbb{P}}\left(1-\frac{1}{p^{s}}\right) \end{align*} $$ where $\mathbb{P}$ is the set of the prime numbers.
EDIT
To derive the continued faction just put $\frac{1}{\zeta(s)}$ in the form $$ \frac{1}{\zeta(s)}=1-\frac{1}{2^s}\left(1+\frac{2^s-1}{3^s}\left(1+\frac{3^s-1}{5^s}\left(1+\frac{5^s-1}{7^s}\left(1+\frac{7^s-1}{11^s}\left(1+\ddots\right ) \right ) \right ) \right ) \right) $$ and then just apply the Euler continued fraction formula. So we can write this as $$ \frac{1}{\zeta(s)}=1-\frac{1}{2^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}\frac{3^s-1}{5^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}\frac{3^s-1}{5^s}\frac{5^s-1}{7^s}-\cdots $$ Now, let $a_1=-\frac{1}{2^s};a_2=\frac{2^s-1}{3^s};a_3=\frac{3^s-1}{5^s};a_4=\frac{5^s-1}{7^s}\cdots$ and we'll get the Euler continued fraction formula.