We need $x u+2 z v+2 y w=1$, $y u+x v+2 z w=0, z u+y v+x w=0$. This is just a straightforward set of linear equations. Solving, we get: $u=\left(x^{2}-2 y z\right) / d, v=\left(2 z^{2}-x y\right) / d, w=$ $\left(y^{2}-x z\right) / d$, where $d=x^{3}+2 y^{3}+4 z^{3}-6 x y z$.

This would fail if $d=0$. But if $d=0$, then multiplying through by a suitable integer we have $6 mnr=m^{3}+2 n^{3}+4 r^{3}$ for some integers $m, n, r$. But we can divide by any common factor of $m, n, r$ to get them without any common factor. But $6 mnr, 2 n^{3}, 4 r^{3}$ are all even, so $m$ must be even. Put $m=2 M$. Then $12 M n r=8 M^{3}+2 n^{3}+4 r^{3}$, so $6 M n r=4 M^{3}+n^{3}+2 r^{3}$. But $6 M n r, 4 M^{3}$ and $2 r^{3}$ are all even, so $n$ must be even. Put $n=2 N$. Then $12 M N r=4 M^{3}+8 N^{3}+2 r^{3}$, so $6 M N r$ $=2 M^{3}+4 N^{3}+r^{3}$, so $r$ must be even. So $m, n, r$ had a common factor 2. Contradiction. So $d$ cannot be zero.