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

$$\int_0^1 \int_0^1 \frac{1}{1+xt+(xt)^2}dxdt=\int_0^1 \int_0^1 \frac{1-xt}{1-(xt)^3}dxdt$$ $$=\sum_{n=0}^\infty \int_0^1 \int_0^1 (xt)^{3n}(1-xt)dxdt=\sum_{n=0}^\infty \int_0^1 \int_0^1 (xt)^{3n}dxdt-\sum_{n=0}^\infty\int_0^1 \int_0^1(xt)^{3n+1}dxdt$$ $$=\sum_{n=0}^\infty \frac{1}{(3n+1)^2}-\sum_{n=0}^\infty \frac{1}{(3n+2)^2}=\frac{\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)}{9}=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{2\pi}{3}\right)$$ Above follows using the relationship between the trigamma function and the Clausen function.

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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.

\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

$$\boxed{I=\frac29\psi_1(1/3)-\frac4{27}\pi^2}$$

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