Answer
$$\frac{7}{145 \sqrt{2}}$$
Work Step by Step
Given
$$
f(x, y)=\tan ^{-1}(x y), \quad \mathbf{v}=\langle 1,1\rangle, \quad P=(3,4)
$$
Since the direction of $\mathbf{v}$ is
\begin{align*}
\mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\
&=\frac{\langle 1,1\rangle}{\sqrt{1+1}}\\
&=\left\langle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\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\frac{y}{1+(x y)^{2}}, \frac{x}{1+(x y)^{2}}\right\rangle \\
\nabla f_{(3,4)}&=\left\langle \frac{4}{145}, \frac{3}{145}\right\rangle
\end{align*}
Then the directional derivatives are given by
\begin{align*}
D_{\mathbf{u}} f(3,4)&=\nabla f_{(3,4)} \cdot \mathbf{u}\\
&=\left\langle \frac{4}{145}, \frac{3}{145}\right\rangle \cdot\left\langle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle\\
&=\frac{7}{145 \sqrt{2}}
\end{align*}
