Answer
$$\frac{5}{3}$$
Work Step by Step
Given
$$
g(x, y, z)=z^{2}-x y^{2}, \quad \mathbf{v}=\langle- 1,2,2\rangle, \quad P=(2,1,3)
$$
Since the direction of $\mathbf{v}$ is
\begin{align*}
\mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\
&=\frac{\langle -1,2,2\rangle}{\sqrt{1+4+4}}\\
&=\left\langle\frac{-1}{3},\frac{2}{3}, \frac{2}{3}\right\rangle
\end{align*}
and
\begin{align*}
\nabla g&=\left\langle\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}
\right\rangle\\
&=\left\langle -y^2, -2xy, 2z\right\rangle \\
\nabla g_{(2,1,3)}&=\left\langle -1, -4,6\right\rangle
\end{align*}
Then the directional derivatives are given by
\begin{align*}
D_{\mathbf{u}} g(2,1,3)&=\nabla f_{(2,1,3)} \cdot \mathbf{u}\\
&=\left\langle -1, -4,6\right\rangle \cdot\left\langle\frac{-1}{3},\frac{2}{3}, \frac{2}{3}\right\rangle\\
&=\frac{1}{3}-\frac{8}{3}+\frac{12}{3}=\frac{5}{3}
\end{align*}
