Show: $\int_0^1\int_0^1\frac{dx\,dt}{(xt)^2+xt+1}=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{2\pi}{3}\right)$ 
How do you prove that $$\int_0^1\int_0^1\frac{dx\,dt}{(xt)^2+xt+1}=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{2\pi}{3}\right)?$$

My proof: 
Suppose we have a sequence $(q_n)_{n\ge0}$ defined by the recurrence $$q_{n+2}=aq_{n+1}+bq_n,\qquad q_0,q_1,a,b \text{ given}.$$
We define the generating function 
$$q(t)=\sum_{n\ge0}q_nt^n,$$
and see that 
$$q(t)-q_1t-q_0=at(q(t)-q_0)+bt^2q(t),$$
so that $$q(t)=\frac{(q_1-aq_0)t+q_0}{1-at-bt^2}.$$
Consider the case 
$$\begin{align}
q_0&=0\\
q_1&=1\\
a&=-1\\
b&=-1,
\end{align}$$
i.e. $q_{n+2}+q_{n+1}+q_n=0$ for all $n\in\Bbb Z_{\ge0}$, with $q_0=0,q_1=1$.
This has the generating function
$$q(t)=\frac{t}{t^2+t+1}.$$
Note that if $q_n=0$ and $q_{n+1}=1$ for any $n\in\Bbb Z_{\ge0}$, then $q_{n+2}=-1$. This implies that $q_n=q_{n+3}$ for all $n$, that is, the sequence is periodic with a period of $3$. The first few values are $0,1,-1,0,1,-1,0,1,-1,..$. Because of the alternating behavior of the sequence, we can represent it as a sine function:
$$q_n=\alpha \sin(n\beta).$$
Since $1=q_1=\alpha\sin\beta\ne 0$ and $-1=q_2=\alpha\sin2\beta\ne0$, both $\alpha$ and $\beta$ are nonzero, and additionally $\beta/\pi\not\in\Bbb Z$. Since $q_n=0$ for $3|n$, $3\beta/\pi\in\Bbb Z$. Finally, if we ensure that $\alpha$ and $\beta$ are both positive, we get that $\beta=2\pi/3$ and $\alpha=2/\sqrt3$. Thus
$$\frac{1}{t^2+t+1}=\frac{2}{\sqrt3}\sum_{n\ge1}\sin\left(\tfrac{2\pi}{3}n\right)t^{n-1}.$$
If we replace $t$ by $xt$ and integrate over $(x,t)\in[0,1]\times[0,1]$, we get
$$\begin{align}
\int_0^1\int_0^1\frac{dxdt}{(xt)^2+xt+1}&=\frac{2}{\sqrt3}\sum_{n\ge1}\sin\left(\tfrac{2\pi}{3}n\right)\int_0^1x^{n-1}dx\int_0^1 t^{n-1}dt\\
&=\frac{2}{\sqrt3}\sum_{n\ge1}\frac{\sin\left(\tfrac{2\pi}{3}n\right)}{n^2}\\
&=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{2\pi}{3}\right).
\end{align}$$
$\square$

From @LeBlanc's comment,
$$\int_0^1 \int_0^1\frac{dx\,dt}{(xt)^2-2xt\cos\theta+1}=\frac1{\sin\theta}\mathrm{Cl}_2(\theta).$$
 A: $$\int_0^1 \int_0^1 \frac{1}{1+xt+(xt)^2}dxdt=\int_0^1 \int_0^1 \frac{1-xt}{1-(xt)^3}dxdt$$
$$=\sum_{n=0}^\infty \int_0^1 \int_0^1  (xt)^{3n}(1-xt)dxdt=\sum_{n=0}^\infty \int_0^1 \int_0^1  (xt)^{3n}dxdt-\sum_{n=0}^\infty\int_0^1 \int_0^1(xt)^{3n+1}dxdt$$
$$=\sum_{n=0}^\infty \frac{1}{(3n+1)^2}-\sum_{n=0}^\infty \frac{1}{(3n+2)^2}=\frac{\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)}{9}=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{2\pi}{3}\right)$$
Above follows using the relationship between the trigamma function and the Clausen function.

An alternative way would be to use the more general generating function:
$$
\frac{\sin t}{1-2x \cos t +x^2}=\sum_{n=1}^\infty x^{n-1} \sin (nt)
$$
Then just set $t=\frac{2\pi}{3}$ and follow the same approach as you did above. A proof for it can be found here.
A: \begin{align}
I&=\int_0^1\int_0^1\frac{dx\ dt}{(xt)^2+xt+1}\overset{xt=y}{=}\int_0^1\int_0^x\frac{dx\ dy}{x(y^2+y+1)}\\
&=\int_0^1\frac{1}{y^2+y+1}\left(\int_y^1\frac{dx}{x}\right)dy=-\int_0^1\frac{\ln y}{y^2+y+1}\ dy\\
&=\int_0^1\frac{y-1}{1-y^3}\ln y\ dy=\int_0^1\frac{y\ln y}{1-y^3}\ dy-\int_0^1\frac{\ln y}{1-y^3}\ dy\\
&=\frac19\int_0^1\frac{x^{-1/3}\ln x}{1-x}\ dx-\frac19\int_0^1\frac{x^{-2/3}\ln x}{1-x}\ dx\\
&=-\frac19\psi_1(2/3)+\frac19\psi_1(1/3)
\end{align}
using $\psi(1-x)-\psi(x)=\pi\cot(\pi x)$, differentiate both sides with respect to $x$ then set $x=2/3$
we get $\psi_1(2/3)=\frac43\pi^2-\psi_1(1/3)$.
Thus
$$\boxed{I=\frac29\psi_1(1/3)-\frac4{27}\pi^2}$$

Note: $\psi_n(z)=-\int_0^1\frac{x^{z-1}\ln^n x}{1-x}\ dx$
