jmc

Dividing the equation by $x^2,$ we get $$6x^2-35x+62-\frac{35}{x}+\frac{6}{x^2}=0.$$Let $y=x+\frac{1}{x}.$ Then $$y^2=x^2+2+\frac{1}{x^2},$$so $x^2+\frac{1}{x^2}=y^2-2.$ Thus, we can re-write the equation above as $$6(y^2-2)-35y+62=0.$$This simplifies to $6y^2-35y+50=0.$ The roots are $y=\frac{5}{2}$ and $y=\frac{10}{3}.$

The roots to $$x+\frac{1}{x}=\frac{5}{2}$$are 2 and $\frac{1}{2}.$ The roots to $$x+\frac{1}{x}=\frac{10}{3}$$are 3 and $\frac{1}{3}.$

Thus, the roots of $6x^4-35x^3+62x^2-35x+6=0$ are $\boxed{2,3,\frac{1}{2},\frac{1}{3}}.$