Any non-zero real numbers $x, y$ such that $y \neq 3$ and $\frac{x}{y}<\frac{x+3}{y-3}$, Will satisfy the condition.

We need to find what condition non-zero real numbers $x, y$ (where $y \neq 3$) must satisfy if:

$\frac{x}{y} < \frac{x+3}{y-3}$

The key insight is to rearrange this inequality and analyze when it holds true.

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Starting with: $\frac{x}{y} < \frac{x+3}{y-3}$

Move everything to one side:

$\frac{x}{y} - \frac{x+3}{y-3} < 0$

Find a common denominator:

$\frac{x(y-3) - y(x+3)}{y(y-3)} < 0$

Expand the numerator:

$\frac{xy - 3x - xy - 3y}{y(y-3)} < 0$

Simplify:

$\frac{-3x - 3y}{y(y-3)} < 0$

Factor out -3:

$\frac{-3(x+y)}{y(y-3)} < 0$

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Since we have $\frac{-3(x+y)}{y(y-3)} < 0$, we can multiply both sides by -1.

When we multiply an inequality by a negative number, the inequality sign flips:

$\frac{3(x+y)}{y(y-3)} > 0$

Since 3 is positive, we can divide both sides by 3:

$\frac{x+y}{y(y-3)} > 0$

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For $\frac{x+y}{y(y-3)} > 0$, we need the numerator and denominator to have the same sign.

This means either:

Both $(x+y) > 0$ and $y(y-3) > 0$, OR

Both $(x+y) < 0$ and $y(y-3) < 0$

Let's analyze $y(y-3)$ in different ranges:

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When $y < 0$:

$y < 0$ (negative)

$y - 3 < -3 < 0$ (negative)

Therefore: $y(y-3) = (\text{negative}) \times (\text{negative}) = \text{positive}$

For $\frac{x+y}{y(y-3)} > 0$, we need $(x+y) > 0$

Condition: When $y < 0$, then $x + y > 0$

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When $0 < y < 3$:

$y > 0$ (positive)

$y - 3 < 0$ (negative)

Therefore: $y(y-3) = (\text{positive}) \times (\text{negative}) = \text{negative}$

For $\frac{x+y}{y(y-3)} > 0$, we need $(x+y) < 0$

Condition: When $0 < y < 3$, then $x + y < 0$

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When $y > 3$:

$y > 0$ (positive)

$y - 3 > 0$ (positive)

Therefore: $y(y-3) = (\text{positive}) \times (\text{positive}) = \text{positive}$

For $\frac{x+y}{y(y-3)} > 0$, we need $(x+y) > 0$

Condition: When $y > 3$, then $x + y > 0$

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The condition that $x, y$ must satisfy is:

$\frac{x+y}{y(y-3)} > 0$

This translates to:

If $y < 0$: then $x + y > 0$

If $0 < y < 3$: then $x + y < 0$

If $y > 3$: then $x + y > 0$

The sign of $(x+y)$ depends on which interval $y$ falls into, and this relationship ensures the original inequality holds.