PROBLEM \# 2: Consider the function $\quad f(x, y, z)$ or $f(x, y)$
a. Find the directional derivative of $f(x, y, z)$ or $f(x, y)$ at the point $P$ in the direction toward the point $Q$ or in the direction of vector $v$.
b. Find the direction in which the function changes most rapidly at the point $P$.
c. Find the maximum rate of change of function at the point $P$.
Suggested problems:
(i) $f(x, y, z)=x-y^{3}-2 z^{2}-2$
$P=(-4,-2,1)$
$v=\langle 3,6,-2\rangle$

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 \(v=\langle 3,6,-2\rangle\).