Answer
$\pm (\dfrac{1}{\sqrt {2}} i+\dfrac{1}{\sqrt {2}} j)$
Work Step by Step
$ f_x= \dfrac{2x(x^2+y^2)-2x(x^2-y^2)}{(x^2+y^2)^2} $ and $ f_y=\dfrac{-2y(x^2+y^2)-2y(x^2-y^2)}{(x^2+y^2)^2} $
$\nabla f =\dfrac{4xy^2}{(x^2+y^2)}i-\dfrac{4yx^2}{(x^2+y^2)}j$
and $\nabla f (1,1)=\dfrac{4(1)(1)^2}{(1^2+1^2)}i-\dfrac{4(1)(1)^2}{(1^2+1^2)}j=i-j$
We know that $D_u f = \nabla f \cdot u$
Now, the unit vector of the gradient vector can be found as:
$ u =\dfrac{1}{\sqrt {1^2+(-1)^2}} i+\dfrac{1}{\sqrt {1^2+(-1)^2}} j$
or, $=\dfrac{1}{\sqrt {2}} i+\dfrac{1}{\sqrt {2}} j$
Thus, $u=\pm (\dfrac{1}{\sqrt {2}} i+\dfrac{1}{\sqrt {2}} j)$
