 Solution

The free part of the abelianization is generated by $[y]$, so the fibration class sends $y\mapsto 1$ and $x\mapsto 0$. Applying Fox calculus to the relator and then substituting $x=1$, $y=t$ givesThis has breadth two. Since $K$ is assumed fibered, the breadth is twice the genus, so its fiber is a genus-one surface with one boundary component:

The monodromy can also be read directly from the relator. Put $x_i=y^{i}x{y}^{-i}$. Reidemeister rewriting givesThus the kernel is freely generated by $(x_{-1},x_0)$, and conjugation by $y$ sendsOn first homology, in the ordered basis $([x_{-1}],[x_0])$, a representative isIts characteristic polynomial is $t^2-4t+1$, agreeing up to the unit $-t$ with $4-t-t^{-1}$.