Answer
$$\frac{6}{ \sqrt{13}}$$
Work Step by Step
Given
$$
f(x, y)=\ln \left(x^{2}+y^{2}\right), \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j}, \quad P=(1,0)
$$
Since the direction of $\mathbf{v}$ is
\begin{align*}
\mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\
&=\frac{\langle 3,-2\rangle}{\sqrt{9+4}}\\
&=\left\langle\frac{3}{\sqrt{13}},\frac{-2}{\sqrt{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\frac{2 x}{x^{2}+y^{2}}, \frac{2 y}{x^{2}+y^{2}}\right\rangle \\
\nabla f_{(1,0)}&=\left\langle 2,0\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 2,0\right\rangle \cdot\left\langle\frac{3}{\sqrt{13}},\frac{-2}{\sqrt{13}}\right\rangle\\
&=\frac{6}{ \sqrt{13}}
\end{align*}
