1. **State the problem:** Find the point on the curve $y=\sqrt{x}$ where the tangent line is parallel to the line $y=\frac{1}{8}x$. 2. **Identify the slope of the given line:** The line $y=\frac{1}{8}x$ has slope $m=\frac{1}{8}$. 3. **Find the derivative of the curve:** The curve is $y=\sqrt{x}=x^{\frac{1}{2}}$. The derivative is $$y' = \frac{d}{dx} x^{\frac{1}{2}} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}.$$ 4. **Set the derivative equal to the slope of the line:** We want the tangent slope to be $\frac{1}{8}$, so $$\frac{1}{2\sqrt{x}} = \frac{1}{8}.$$ 5. **Solve for $x$: ** Multiply both sides by $2\sqrt{x}$: $$\cancel{2\sqrt{x}} \times \frac{1}{\cancel{2\sqrt{x}}} = \frac{1}{8} \times 2\sqrt{x} \implies 1 = \frac{2\sqrt{x}}{8} = \frac{\sqrt{x}}{4}.$$ Multiply both sides by 4: $$4 = \sqrt{x}.$$ Square both sides: $$4^2 = (\sqrt{x})^2 \implies 16 = x.$$ 6. **Find the corresponding $y$ value:** $$y = \sqrt{16} = 4.$$ 7. **Final answer:** The point on the curve where the tangent line is parallel to $y=\frac{1}{8}x$ is $$\boxed{(16, 4)}.$$