jmc

Note using the change-of-base formula that ${\log }_{y}x{\log }_{x}y=1$. We find that $$\begin{array}{rl}({\log }_{y}x{)}^2+({\log }_{x}y{)}^2& =({\log }_{y}x{)}^2+2{\log }_{y}x{\log }_{x}y+({\log }_{x}y{)}^2-2{\log }_{y}x{\log }_{x}y\\ & =({\log }_{y}x+{\log }_{x}y{)}^2-2{\log }_{y}x{\log }_{x}y\\ & =7^2-2\\ & =\boxed{47}.\end{array}$$