Let $F$ be the free group on two elements $x$, $y$. To make our lives easier, we denote $x^{-1}$ by $X$ and $y^{-1}$ by $Y$. Let $g_i$ for $i = 1, \ldots, 6$ be the following elements:$$g_1 = xyxY,$$$$g_2 = XYXyxYxyXyxYXYXyxYxyxyXYY,$$$$g_3 = XYXyx,$$$$g_4 = YxyXyxYXYXyyx,$$$$g_5 = YxyXYxyxyX,$$$$g_6 = xxYXXy.$$Let $G$ be the subgroup of $F$ generated by the $g_i$.
Question. How do I see that $G$ is free of rank $6$?
I'm not an expert in free goups, but I'll give it a go: simplifying the generators like @Derek suggested:
$$g_1 = x\color{red}{yx}Y,$$ $$g_2 = XYx\color{red}{YxyxyXY}Y,$$ $$g_3 = XY\color{red}{X}yx,$$ $$g_4 = YxyX\color{red}{yxYXYXy}yx,$$ $$g_5 = YxyX\color{red}{Yxyxy}X,$$ $$g_6 = x\color{red}{xYX}Xy.$$ $$G_1 = y\color{red}{XY}X,$$ $$G_2 = y\color{red}{yxYXYXy}Xyx,$$ $$G_3 = XY\color{red}{x}yx,$$ $$G_4 = XY\color{red}{YxyxyXY}xYXy,$$ $$G_5 = x\color{red}{YXYXy}xYXy,$$ $$G_6 = Yx\color{red}{xyX}X$$ Now consider all possible concatenations $g_ig_j, G_ig_j, g_iG_j, G_iG_j$ (with $i\neq j$). You will see that the red part of the words never cancel (provided I didn't make a mistake). This implies that in any concatenation of arbitrary length of the $G$'s and $g$'s, the red part will never cancel. This means that there are no relations possible between them, and they indeed form a free group of rank $6$.