#### Answer

$$\frac{15}{ \sqrt{14}}$$

#### Work Step by Step

Given
$$
f(x, y)=xy+ z^{3}, \quad P=(3,-2,-1)
$$
Since $$PO= (-3,2,1) $$
Then the direction of $\mathbf{v}$ is
\begin{align*}
\mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\
&=\frac{\langle -3,2,1\rangle}{\sqrt{9+4+1}}\\
&=\left\langle\frac{-3}{\sqrt{14}},\frac{ 2}{\sqrt{14}},\frac{ 1}{\sqrt{14}}\right\rangle
\end{align*}
and
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right\rangle\\
&=\left\langle y ,x, 3z^2\right\rangle \\
\nabla f_{(3,-2,-1)}&=\left\langle -2,3,3\right\rangle
\end{align*}
Then the directional derivatives are given by
\begin{align*}
D_{\mathbf{u}} f(3,-2,-1)&=\nabla f_{(3,-2,-1)} \cdot \mathbf{u}\\
&=\left\langle -2,3,3\right\rangle \cdot\left\langle\frac{-3}{\sqrt{14}},\frac{ 2}{\sqrt{14}},\frac{ 1}{\sqrt{14}}\right\rangle\\
&=\frac{15}{ \sqrt{14}}
\end{align*}