Answer
$$\frac{-50}{ \sqrt{13}}$$
Work Step by Step
Given
$$
f(x, y)=x^{2}+4 y^{2}, \quad P=(3,2)
$$
Since $$Op= (-3,-2) $$
Then 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 2x,8y\right\rangle \\
\nabla f_{(3,2)}&=\left\langle 6,16\right\rangle
\end{align*}
Then the directional derivatives are given by
\begin{align*}
D_{\mathbf{u}} f(3,2)&=\nabla f_{(3,2)} \cdot \mathbf{u}\\
&=\left\langle 6,16\right\rangle \cdot\left\langle\frac{-3}{\sqrt{13}},\frac{-2}{\sqrt{13}}\right\rangle\\
&=\frac{-50}{ \sqrt{13}}
\end{align*}
