Use implicit differentiation to find $y^{\prime}$ and then evaluate $y^{\prime}$ at $(-2,1)$. \[ x^{3}+8 y^{2}=\ln y \] \[ \begin{array}{l} y^{\prime}=\square \\ \left.y^{\prime}\right|_{(-2,1)}=\square \text { (Simplify your answer.) } \end{array} \]

Step 1 ：First, we differentiate both sides of the equation with respect to \(x\).

Step 2 ：The derivative of \(x^{3}\) with respect to \(x\) is \(3x^{2}\).

Step 3 ：The derivative of \(8y^{2}\) with respect to \(x\) is \(16yy'\).

Step 4 ：The derivative of \(\ln y\) with respect to \(x\) is \(\frac{1}{y}y'\).

Step 5 ：So, we have \(3x^{2}+16yy'=\frac{1}{y}y'\).

Step 6 ：Rearrange the equation to solve for \(y'\), we get \(y'=\frac{3x^{2}y}{16y^{2}-1}\).

Step 7 ：Now, we substitute \((-2,1)\) into the equation, we get \(y'=\frac{3(-2)^{2}(1)}{16(1)^{2}-1}\).

Step 8 ：Simplify the equation, we get \(y'=\frac{12}{15}\).

Step 9 ：So, the final answer is \(y'=\frac{4}{5}\).