(d) The asymptotes are $$ y=\frac{\sqrt{2}}{6} x-\frac{7 \sqrt{2}}{6} $$ and $$ y=-\frac{\sqrt{2}}{6} x+\frac{7 \sqrt{2}}{6} $$.

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1. Identify the center of the hyperbola. From the asymptote formulas for a hyperbola, the equations are usually written in the form  
   $$
   y = k \pm \frac{b}{a}(x - h)
   $$  
   where $$(h,k)$$ is the center. Comparing with the given asymptotes  
   $$
   y=\frac{\sqrt{2}}{6} x-\frac{7 \sqrt{2}}{6} \quad \text{and} \quad y=-\frac{\sqrt{2}}{6} x+\frac{7 \sqrt{2}}{6},
   $$  
   we can rewrite them as  
   $$
   y=\frac{\sqrt{2}}{6}(x-7) \quad \text{and} \quad y=-\frac{\sqrt{2}}{6}(x-7).
   $$  
   This shows that the center is at $$(h,k) = (7,0)$$ and the slopes are $$\pm \frac{\sqrt{2}}{6}$$.

2. Recall that for a hyperbola centered at $$(h,k)$$ with horizontal transverse axis, the asymptotes are given by  
   $$
   y = k \pm \frac{b}{a}(x-h).
   $$  
   Here the ratio $$\frac{b}{a}$$ is identified as $$\frac{\sqrt{2}}{6}$$.

3. Substitute the center $$(7,0)$$ and $$\frac{b}{a} = \frac{\sqrt{2}}{6}$$ into the asymptote formulas:  
   $$
   y = 0 \pm \frac{\sqrt{2}}{6}(x-7).
   $$

4. This yields the two asymptote equations:  
   $$
   y = \frac{\sqrt{2}}{6}(x-7) = \frac{\sqrt{2}}{6}x - \frac{7\sqrt{2}}{6}
   $$  
   and  
   $$
   y = -\frac{\sqrt{2}}{6}(x-7) = -\frac{\sqrt{2}}{6}x + \frac{7\sqrt{2}}{6}.
   $$

Thus, the asymptotes are  
$$
y=\frac{\sqrt{2}}{6} x-\frac{7 \sqrt{2}}{6} \quad \text{and} \quad y=-\frac{\sqrt{2}}{6} x+\frac{7 \sqrt{2}}{6}.
$$
The asymptotes are $$ y=\frac{\sqrt{2}}{6} x-\frac{7 \sqrt{2}}{6} $$ and $$ y=-\frac{\sqrt{2}}{6} x+\frac{7 \sqrt{2}}{6} $$.

(d) The asymptotes are and .

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