jmc

We notice that $f(x)$ has only even powers of $x,$ so if we let $y=x^2,$ we can write $$f(x)=15x^4-13x^2+2=15y^2-13y+2=(3y-2)(5y-1).$$Substituting back $x^2$ for $y$ gives us $$f(x)=(3x^2-2)(5x^2-1).$$Then the roots of $f(x)$ are the roots of $3x^2-2$ and $5x^2-1$, which are $\sqrt{\frac{2}{3}},-\sqrt{\frac{2}{3}},\frac{1}{\sqrt{5}},$ and $-\frac{1}{\sqrt{5}}$. Therefore, the greatest root is $\sqrt{\frac{2}{3}}=\boxed{\frac{\sqrt{6}}{3}}.$