Note that $yxy = x^{-1}$, and more generally $yx^ny = (yxy)(yxy)(yxy)\cdots (yxy) = x^{-n}$. ($y$ has order two.) We can use these relations, along with the derived relations $yx^n = x^{-n}y$ (again, $y$ has order two) to shift all the $x$ to the left (or right, if you want, depending on your definition of "canonical expression") of some expression. For example:
The first one is equivalent to $x^{11} y x^4 y^3 x^{-2}$. Applying the relation $yx^4y = x^{-4}$ we get that this is equivalent to $x^{11}x^{-4}y^2x^{-2} = x^7 y^2 x^{-2}$. Applying the relation $yx^{-2} = x^2 y$ to this we get that this is equivalent to $x^7 yx^2 y$. Finally, applying $yx^2 y = x^{-2}$ we find that the original expression is just $x^7 x^{-2} = x^5$.