Show this $\int_0^\infty \frac{t\ln(2\sinh t)}{\left(3t^2+\ln^2(2\sinh t)\right)^2}~dt=0$

This answer directly proves that:

$$\text{Im}\int_0^\infty \frac{dt}{\left(i t\sqrt{3}+\ln(2\sinh t)\right)^2}=0$$

First, we make a change of variable:

$$x=e^{-2t}$$

Which transforms the identity to:

$$\text{Im} \int_0^1 \frac{dx}{x \left(\ln(1-x)-e^{\pi i/3} \ln x \right)^2}=0$$

Finding the imaginary part explicitly, we now need to prove:

$$ \int_0^1 \frac{\ln x \ln(1-x)-\frac{1}{2} \ln^2 x}{x \left(\ln^2 x+\ln^2(1-x)-\ln x \ln (1-x)\right)^2}dx=0$$

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Let's introduce a function:

$$f(x)=f(1-x)=\ln^2 x+\ln^2(1-x)-\ln x \ln (1-x)$$

As we have a difference of two positive definite functions under the integral, the identity is equivalent to:

$$\int_0^1 \frac{\ln x \ln(1-x)}{x f(x)^2}dx=\frac{1}{2}\int_0^1 \frac{ \ln^2 x}{x f(x)^2}dx$$

Let's denote the integrals $J_1$ and $J_2$. We need to prove that $J_1=J_2$.

Using the substitution $x \to 1-x$ we can prove the following identities:

$$J_1=\int_0^1 \frac{\ln x \ln(1-x)}{(1-x) f(x)^2}dx=\frac{1}{2} \int_0^1 \frac{\ln x \ln(1-x)}{x (1-x) f(x)^2}dx$$

$$J_2=\frac{1}{2}\int_0^1 \frac{ \ln^2 (1-x)}{(1-x) f(x)^2}dx=\frac{1}{4}\int_0^1 \frac{ (1-x)\ln^2 x+x\ln^2 (1-x)}{x(1-x) f(x)^2}dx$$

Subtracting the two forms of $J_2$ gives us another set of identities:

$$J_3=\int_0^1 \frac{ \ln^2 x}{x (1-x)f(x)^2}dx=\int_0^1 \frac{ \ln^2 (1-x)}{x (1-x)f(x)^2}dx=\int_0^1 \frac{ \ln^2 x+\ln^2 (1-x)}{x f(x)^2}dx$$

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From the above follows a relation:

$$J_3-J_1=\int_0^1 \frac{ 1}{x f(x)}dx$$

Now we use integration by parts with:

$$u(x)=\frac{ 1}{f(x)}, \qquad v(x)= \ln x$$

The limits for $u(x)v(x)$ at $0$ and $1$ are both equal to zero. After simplifications, we can write:

$$J_3-J_1=\int_0^1 \frac{(2-x) \ln^2 x-(1+x) \ln x \ln (1-x)}{x(1-x)f(x)^2}dx$$

Making a substitution $x \to 1-x$ and adding the two results, we obtain a symmetric form of the integral:

$$J_3-J_1=\frac{1}{2} \int_0^1 \frac{(2-x) \ln^2 x+(1+x) \ln^2 (1-x)-3 \ln x \ln (1-x)}{x(1-x)f(x)^2}dx$$

From the other identities above, it can be finally seen that:

$$J_3-J_1=J_3+2J_2-3J_1$$

Or immediately:

$$J_1=J_2$$

The proof is finished.

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Remark. This doesn't use the other identity shown by FDP in the comments. To me it looks very difficult to prove.