Step 1 :Given the function \(f(x, y, z)=x-y^{3}-2 z^{2}-2\), the point \(P=(-4,-2,1)\), and the vector \(v=\langle 3,6,-2\rangle\).
Step 2 :First, we need to find the partial derivatives of the function with respect to \(x\), \(y\), and \(z\).
Step 3 :The partial derivative of \(f\) with respect to \(x\) is \(\frac{df}{dx} = 1\).
Step 4 :The partial derivative of \(f\) with respect to \(y\) is \(\frac{df}{dy} = -3y^{2}\).
Step 5 :The partial derivative of \(f\) with respect to \(z\) is \(\frac{df}{dz} = -4z\).
Step 6 :Next, we evaluate these partial derivatives at the point \(P=(-4,-2,1)\).
Step 7 :\(\frac{df}{dx} = 1\) at \(P\).
Step 8 :\(\frac{df}{dy} = -12\) at \(P\).
Step 9 :\(\frac{df}{dz} = -4\) at \(P\).
Step 10 :Then, we find the unit vector in the direction of \(v\). The magnitude of \(v\) is \(7\), so the unit vector is \(\langle \frac{3}{7}, \frac{6}{7}, -\frac{2}{7} \rangle\).
Step 11 :Finally, we calculate the dot product of the gradient and the unit vector to find the directional derivative. The directional derivative is \(-\frac{61}{7}\).
Step 12 :\(\boxed{-\frac{61}{7}}\) is the directional derivative of the function \(f(x, y, z)=x-y^{3}-2 z^{2}-2\) at the point \(P=(-4,-2,1)\) in the direction of the vector \(v=\langle 3,6,-2\rangle\).