Let $f(x, y, z)=8 x y-z^{2}, x=4 r \cos (\theta), y=\cos ^{2}(\theta), z=3 r$. Use the Chain Rule to calculate the partial derivative. (Use symbolic notation and fractions where needed. Express the answer in terms of independent variables.) \[ \frac{\partial f}{\partial \theta}= \]

Step 1 ：Express the function \(f\) in terms of \(r\) and \(\theta\) by substituting \(x=4r\cos(\theta)\), \(y=\cos^2(\theta)\), and \(z=3r\) into \(f(x, y, z)\). This gives us: \[f(r, \theta) = 8(4r\cos(\theta))(\cos^2(\theta)) - (3r)^2\] which simplifies to \[f(r, \theta) = 32r\cos^3(\theta) - 9r^2\]

Step 2 ：Calculate the partial derivative of \(f\) with respect to \(\theta\): \[\frac{\partial f}{\partial \theta} = \frac{\partial}{\partial \theta}(32r\cos^3(\theta) - 9r^2)\]

Step 3 ：Use the chain rule to find the derivative: \[\frac{\partial f}{\partial \theta} = 32r \cdot 3\cos^2(\theta) \cdot -\sin(\theta) - 0\] which simplifies to \[\frac{\partial f}{\partial \theta} = -96r\cos^2(\theta)\sin(\theta)\]

Step 4 ：So, the partial derivative of \(f\) with respect to \(\theta\) is \(\boxed{-96r\cos^2(\theta)\sin(\theta)}\)