Answer
$$\frac{34}{13}$$
Work Step by Step
Given
$$
f(x, y)=e^{x y-y^{2}}, \quad \mathbf{v}=\langle 12,-5\rangle, \quad P=(2,2)
$$
Since the direction of $\mathbf{v}$ is
\begin{align*}
\mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\
&=\frac{\langle 12,-5\rangle}{\sqrt{144+25}}\\
&=\left\langle\frac{12}{13},\frac{-5}{13}\right\rangle
\end{align*}
and
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\
&=\left\langle e^{xy-y^2}y,e^{xy-y^2}\left(x-2y\right)\right\rangle \\
\nabla f_{(2,2)}&=\left\langle 2,-2\right\rangle
\end{align*}
Then the directional derivatives are given by
\begin{align*}
D_{\mathbf{u}} f(2,2)&=\nabla f_{(2,2)} \cdot \mathbf{u}\\
&=\left\langle 2,-2\right\rangle \cdot\left\langle\frac{12}{13},\frac{-5}{13}\right\rangle\\
&=\frac{24+10}{13}= \frac{34}{13}
\end{align*}
