Answer
$u=\pm (\dfrac{2}{\sqrt {53}} i+\dfrac{7}{\sqrt {53}} j)$
Work Step by Step
Since, $D_u f = \nabla f \cdot u$
$ f_x (3,2) = y=2 \\ f_y (3,2)= x+2y =3+(2)(2)=3+4=7$
$\implies \nabla f =2 i+7j$
The unit vector of the gradient vector is
$ \nabla f =\dfrac{2 i}{\sqrt {2^2+7^2}}+\dfrac{7 j}{\sqrt {2^2+7^2}} \\ =\dfrac{2}{\sqrt {53}} i+\dfrac{7}{\sqrt {53}} j $
and $u=\pm (\dfrac{2}{\sqrt {53}} i+\dfrac{7}{\sqrt {53}} j)$
