$\frac{\tan^2(3x)}{1+\tan^2(3x)}+\frac{\tan^2(2x)}{1+\tan^2(2x)}+\frac{\tan^2(x)}{1+\tan^2(x)}=2$ has no solution?

$$\frac{\tan^2(3x)}{1+\tan^2(3x)}+\frac{\tan^2(2x)}{1+\tan^2(2x)}+\frac{\tan^2(x)}{1+\tan^2(x)}=2$$ if we simplify ,we have $$\sin^2(3x)+\sin^2(2x)+\sin^2(x)=2\\\frac{1-\cos(6x)}{2}+\frac{1-\cos(4x)}{2}+\frac{1-\cos(2x)}{2}=2\\ \cos(6x)+\cos(4x)+\cos(2x)=-1\\\cos(6x)+\cos(2x)=-(1+\cos(4x))\\2\cos(4x)\cos(2x)=-2\cos^2(2x)\\\begin{cases}\cos(2x)= 0 \to 2x=\pm\frac{\pi}{2}+k\pi\\\cos(2x)=-\cos(4x)\to \begin{cases}\cos(2x)=-1\\\cos(2x)=\frac12\end{cases}\end{cases},$$ but all of the roots are not acceptable because ,the denominator of $\tan(3x)$ or $\tan(2x)$ or$\tan(x)$ going to be zero .
Is my work true ? If draw the $f(x)=\frac{\tan^2(3x)}{1+\tan^2(3x)}+\frac{\tan^2(2x)}{1+\tan^2(2x)}+\frac{\tan^2(x)}{1+\tan^2(x)}$ and $g(x)=2$ by desmos ...we will see some roots $\frac{\pi}{6} ,\frac{\pi}{2}$ are recognizable ,but what about the other ?
https://www.desmos.com/calculator/x9ikgkbyhp

I think you are right and our equation has no solutions.

"Roots" $a$ which you see they are because $$\lim_{x\rightarrow a}\left(\frac{\tan^23x}{1+\tan^23x}+\frac{\tan^22x}{1+\tan^22x}+\frac{\tan^2x}{1+\tan^2x}\right)=2.$$ It follows from your solution, but the domain says that they are not roots.

The problem here is that whoever has produced this problem has constructed a form for $sin^2(x)$ which is artificial for the purpose of the puzzle.

It therefore seems like there is a problem.

But there is not because the everywhere in $\mathbb{R}$ the expression involving $tan(x)$ is always going to reduce to $sin^2(x)$.

The artificiality of using $tan(x)$ is a red herring.

It would be like defining $x$ as $\frac {x^2}{x}$ and then claiming there was a singular point at $x=0$ (which there is not).

Suppose otherwise there exists $x\in{\bf R}$ such that $$\frac{\tan^2(3x)}{1+\tan^2(3x)}+\frac{\tan^2(2x)}{1+\tan^2(2x)}+\frac{\tan^2(x)}{1+\tan^2(x)}=2.$$ If we simplify (using the formula $\sin^2a+\cos^2a=1$ for each term), we have $$\sin^2(3x)+\sin^2(2x)+\sin^2(x)=2\\ \Rightarrow \frac{1-\cos(6x)}{2}+\frac{1-\cos(4x)}{2}+\frac{1-\cos(2x)}{2}=2\\ \Rightarrow \cos(6x)+\cos(4x)+\cos(2x)=-1\\ \Rightarrow\cos(6x)+\cos(2x)=-(1+\cos(4x))$$ which implies that (this step is too quick for me, I don't see why (*) is obviously true.) $$2\cos(4x)\cos(2x)=-2\cos^2(2x)\tag{*}$$ $$\tag{**} \begin{cases} \cos(2x)= 0 \to 2x=\pm\frac{\pi}{2}+k\pi\\ \cos(2x)=-\cos(4x)\to \begin{cases}\cos(2x)=-1\\ \cos(2x)=\frac12 \end{cases} \end{cases},$$ (Too quick. Why does $\cos(2x)=-\cos(4x)$ imply the values you claim in (**)?) but all of the roots are not acceptable because, the denominator of $\tan(3x)$ or $\tan(2x)$ or $\tan(x)$ going to be zero. ($\cos(2x)=0$ certainly gives a contradiction. How do $-1$ or $1/2$ do so?)

Is my work true? (Unclear. You don't have obvious logic mistakes but proof for several important places is missing.)