Fixed Field of $\sigma, \tau$

You can get a complete answer by checking out the related answer that @JyrkiLahtonen mentions in his comment. But I see this as not so much an algebraic question, as one in complex variable theory and geometry.

Do you know the relationship between $2\times2$ complex matrices and fractional-linear transformations of the extended complex plane, also known as the Riemann sphere? You get a homomorphism from nonsingular matrices $\pmatrix{a&b\\c&d}$ to transformations $z\mapsto\frac{az+b}{cz+d}$ of the Riemann sphere. The kernel is the group of scalar matrices $aI$, where $I$ is the identity matrix. Then your $\sigma$ is $z\mapsto1/z$, coming from $\pmatrix{0&1\\1&0}$ and your $\tau$ comes from $\pmatrix{-1&1\\0&1}$. The product of these in either order gives you something of order three.

You check that your six transformations permute the subset $\{0,1,\infty\}$ of the Riemann sphere, and so it makes sense \, if you want something fixed by all six, to try a rational function $h(z)$ with poles at $0$, $1$,and infinity. In fact, it’s a theorem in algebraic curves that the field extension degree $[\mathbb C(z)\colon\mathbb C(h(z))]$ is the total number of poles of $h$, counting multiplicity. So you should expect that if my strategy works, $h$ will have double poles at the three points I mentioned.

If we put $z^2(z-1)^2$ in the denominator of $h$, and look for a sextic numerator, then there will also be a double pole at infinity. What about special points of your two given transformations? If you look at $1/2$, you see that it’s fixed under $\tau$, but sent to $2$ by $\sigma$. Then you check that the set $\{-1,1/2,2\}$ is invariant under both $\sigma$ and $\tau$. So why not choose $$ h=\frac{(z+1)^2(2z-1)^2(z-2)^2}{z^2(z-1)^2}\,? $$ There it is, sextic above, obviously invariant under $\tau$, and almost as obviously invariant under $\sigma$.

EDIT: Here’s my response to your request for a description of how to find an irreducible sextic, showing that you readyy do have a generator of the subfield $F$. You start with your rational function known to be fixed under $\sigma$ and $\tau$. It will be a suitable $f(x)/g(x)$, and the max of the degrees of $f$ and $g$ will be six. All our computed examples have $g$ of smaller degree, and for my explanation, I will assume that this is the case, but the general situation is only a little more complicated.

We have $u=f(x)/g(x)$, fixed under the order-six group $G$ generated by $\sigma$ and $\tau$. We know, from general Galois-theoretical considerations, that the fixed field $F$ has $[k(x)\colon F]=6$. Since $u\in F$, we know that $[k(x)\colon k(u)]\ge6$. On the other hand consider the sextic polynomial $f(T)-g(T)u\in k(u)[T]$. I have used $\deg(g)<\deg(f)=6$ to conclude that this polynomial in $T$ is of degree six. Now, $x$ is clearly a root of our polynomial, so $[(k(u))(x)\colon k(u)]\le 6$; but of course $(k(u))(x)=k(x)$, so that $[k(x)\colon k(u)]=6$, and hence $F=k(u)$. You get as a bonus the conclusion that our polynomial in $T$ is irreducible over $k(u)$.