1. **State the problem:** We want to find the values of $r$ such that the function $y = e^{r x}$ satisfies the differential equation $$\frac{d^2 y}{dx^2} - 6 \frac{dy}{dx} + 8 y = 0.$$\n\n2. **Write down the derivatives:**\n- First derivative: $$\frac{dy}{dx} = r e^{r x}.$$\n- Second derivative: $$\frac{d^2 y}{dx^2} = r^2 e^{r x}.$$\n\n3. **Substitute into the differential equation:**\n$$r^2 e^{r x} - 6 r e^{r x} + 8 e^{r x} = 0.$$\n\n4. **Factor out $e^{r x}$ (which is never zero):**\n$$e^{r x} (r^2 - 6 r + 8) = 0.$$\nSince $e^{r x} \neq 0$, we have\n$$r^2 - 6 r + 8 = 0.$$\n\n5. **Solve the quadratic equation:**\n$$r^2 - 6 r + 8 = 0.$$\nUse the quadratic formula: $$r = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm \sqrt{4}}{2}.$$\n\n6. **Simplify the roots:**\n$$r = \frac{6 \pm 2}{2}.$$\nThis gives two values:\n- $$r = \frac{6 + 2}{2} = \frac{8}{2} = 4,$$\n- $$r = \frac{6 - 2}{2} = \frac{4}{2} = 2.$$\n\n**Final answer:** The values of $r$ are $2$ and $4$.