Is it mathematically possible to solve this problem?

We have

$$x_i=sqrt[3]{x_1x_2x_3},e^{y_i}.$$

Multiplying the equations for $i=1,2,3$, you get the compatibility condition

$$1=e^{y_1+y_2+y_3}.$$ Without it, no solution exists.

Then using ratios,

$$x_2=x_1e^{y_2-y_1}$$ $$x_3=x_1e^{y_3-y_1}$$ and you are free to choose $x_1$.

I think what you are asking is that for vectors: $$y=begin{pmatrix}y_1\y_2\y_3end{pmatrix},x=begin{pmatrix}x_1\x_2\x_3end{pmatrix}$$ then is there a way to calculate: $$y=begin{pmatrix}logfrac{x_1}{(prod x)^{1/3}}\logfrac{x_2}{(prod x)^{1/3}}\logfrac{x_3}{(prod x)^{1/3}}end{pmatrix}$$ notice that: $$y_1+y_2+y_3=log(x_1x_2x_3)-3log(prod x)^{1/3}=log(prod x)-log(prod x)=0$$ and since we know $|y|=8$ we have the system: $$y_1+y_2+y_3=0$$ $$y_1^2+y_2^2+y_3^2=64$$