How to divide two inequalities

Suppose $x^2-y^2 < 8$ and $x+y > 3$. If $y - x \le -\dfrac{8}{3}$, then $x-y \ge \dfrac{8}{3}$, and hence $x^2-y^2 = (x-y)(x+y) > \dfrac{8}{3} \cdot 3 = 8$, a contradiction since $x^2-y^2 < 8$.

Thus, any $(x,y)$ which satisfy both $x^2-y^2 < 8$ and $x+y > 3$ will also satisfy $y - x < -\dfrac{8}{3}$.

However, the converse is not true, that is if $(x,y)$ satisfy $y-x < -\dfrac{8}{3}$, then it is not necessarily true that $x^2-y^2 < 8$ and $x+y > 3$. See the other comments for counterexamples.

Following the comment above, here are some small steps:

Let us graph the two inequalities (using a domain where the corresponding functions exist).

The first one holds when $y^2 > x^2 - 8$.

So we can graph $y = \sqrt{x^2 - 8}$ and color the points above it blue; but, since there are two square roots, we also consider $y = -\sqrt{x^2 - 8}$ and color the points below it yellow.

Next, graph the second inequality, $y > -x + 3$, by drawing the line $y = -x + 3$, and color the points above it red.

Now, anything that is colored purple (i.e., red and blue) or orange (i.e., red and yellow) satisfies the inequalities.

Does this picture correspond to your conjectured $y > x - 8/3$?