Square both sides:
Move $8+2x^2+2y^2$ to the RHS and square again:
$$4(16 + 8 x^2 + x^4 - 8 y^2 + 2 x^2 y^2 + y^4)=784 - 112 x^2 + 4 x^4 - 112 y^2 + 8 x^2 y^2 + 4 y^4$$
The second problem is similar.
If you wanted to "cheat" or check your work, you could use the fact that in an ellipse, $r_1 + r_2 = 2a$, in your case $r_1 + r_2 = 6 \to a = 3$, and also the distance between focii is $4 = 2c$. Thus you could have found the coefficient of $y^2$ as $1/a^2=1/9$, and that of $x^2$ as $1/(a^2-c^2)=1/5$