Answer
$$\frac{39}{4 \sqrt{2}}$$
Work Step by Step
Given
$$
f(x, y)=x^{2} y^{3}, \quad \mathbf{v}=\mathbf{i}+\mathbf{j}, \quad P=\left(\frac{1}{6}, 3\right)
$$
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 2 xy^3, 3x y^{2}\right\rangle \\
\nabla f_{(1/6,3)}&=\left\langle 9, \frac{3}{4}\right\rangle
\end{align*}
Then the directional derivatives are given by
\begin{align*}
D_{\mathbf{u}} f(1/6,3)&=\nabla f_{(1/6,3)} \cdot \mathbf{u}\\
&=\left\langle 9, \frac{3}{4}\right\rangle \cdot\left\langle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle\\
&=\frac{9}{\sqrt{2}}+\frac{3}{4 \sqrt{2}}=\frac{39}{4 \sqrt{2}}
\end{align*}
