#### Answer

$$\frac{44}{5} $$

#### Work Step by Step

Given
$$
f(x, y)=x^{2}+y^{3}, \quad \mathbf{v}=\langle 4,3\rangle, \quad P=(1,2)
$$
Find the direction vector:
\begin{align*}
\mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\
&=\frac{\langle 4,3\rangle}{\sqrt{16+9}}\\
&=\left\langle\frac{4}{5}, \frac{3}{5}\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 x, 3 y^{2}\right\rangle \\
\nabla f_{(1,2)}&=\langle 2,12\rangle
\end{align*}
Then the directional derivative is given by
\begin{align*}
D_{\mathbf{u}} f(1,2)&=\nabla f_{(1,2)} \cdot \mathbf{u}\\
&=\langle 2,12\rangle \cdot\left\langle\frac{4}{5}, \frac{3}{5}\right\rangle\\
&=\frac{8}{5}+\frac{36}{5}\\
&=\frac{44}{5}
\end{align*}