Find a function $y(x)$ such that $8 y y^{\prime}=x \quad$ and $\quad y(8)=3$.
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Step 1 ：Rewrite the given differential equation as \(y' = \frac{x}{8y}\).

Step 2 ：Separate the variables and integrate: \(\int y dy = \int \frac{x}{8} dx\).

Step 3 ：This gives us: \(\frac{1}{2} y^2 = \frac{1}{16} x^2 + C\).

Step 4 ：Multiply through by 2 to simplify: \(y^2 = \frac{1}{8} x^2 + 2C\).

Step 5 ：Use the initial condition \(y(8) = 3\) to solve for \(C\). Substituting \(x = 8\) and \(y = 3\) into the equation gives: \(9 = 8 + 2C\).

Step 6 ：Solve for \(C\): \(2C = 1\) so \(C = \frac{1}{2}\).

Step 7 ：Substitute \(C\) back into the equation: \(y^2 = \frac{1}{8} x^2 + 1\).

Step 8 ：Take the square root of both sides to get two possible solutions: \(y = \sqrt{\frac{1}{8} x^2 + 1}\) or \(y = -\sqrt{\frac{1}{8} x^2 + 1}\).

Step 9 ：Since \(y(8) = 3\) is positive, we choose the positive root.

Step 10 ：\(\boxed{y = \sqrt{\frac{1}{8} x^2 + 1}}\) is the solution to the differential equation that satisfies the given initial condition.