Find the minimium point of a function $f(x, y)$
You can complete the square and write $$(f(x,y))^2=42\Bigg(x+\frac{2}{21}y+\frac{41}{42}\Bigg)^2+{223 \over 21}\Bigg(y+\frac{464}{223}\Bigg)^2+{1 \over 446}$$ You should now see that $f(x,y)$ attains a minimum value of $\frac{1}{\sqrt{446}}$ at $(x,y)=\Big(-\frac{347}{446},-\frac{464}{223}\Big)$.