Answer
$u_{max}= \lt 0,1 \gt$
and $u_{min}= \lt 0,-1 \gt$
$D_u f_{max} = 2$
and $D_u f_{min} =-2$
Work Step by Step
$ f_x= 2xy+ye^{xy} \sin y \implies f_x(1,0)= 0$ and $ f_y= x^2+e^{xy} \cos y +x e^{xy} \sin y \implies f_y(1,0)=(1)^2+e^{(1)(0)} \cos (0) +(1) e^{(1)(0)} \sin (0)=2$
The directions of the unit vector are as follows:
$u_{max}=\dfrac{\nabla f (,0)}{|\nabla f (1,0) |}= \lt 0,1 \gt$
and $u_{min}=- u_{max}= \lt 0,-1 \gt$
We know that $D_u f = \nabla f \cdot u$
Therefore, $D_u f_{max} =\nabla f (1,0) \cdot u_{max} =\lt 0,2 \gt \cdot \lt 0,1 \gt = 2$
and $D_u f_{min} =-D_u f_{max} =-2$
