## Thomas' Calculus 13th Edition

$$u=\pm (\dfrac{1}{\sqrt {2}} i+\dfrac{1}{\sqrt {2}} j)$$
Since, $D_u f = \nabla f \cdot u$ $$f_x= \dfrac{2x(x^2+y^2)-2x(x^2-y^2)}{(x^2+y^2)^2} \\ f_y=\dfrac{-2y(x^2+y^2)-2y(x^2-y^2)}{(x^2+y^2)^2}$$ $\nabla f (1,1) =\dfrac{4xy^2}{(x^2+y^2)}i-\dfrac{4yx^2}{(x^2+y^2)}j = \dfrac{4(1)(1)^2}{(1^2+1^2)}i-\dfrac{4(1)(1)^2}{(1^2+1^2)}j=i-j$ The unit vector of a gradient vector is: $$u =\dfrac{1}{\sqrt {1^2+(-1)^2}} i+\dfrac{1}{\sqrt {1^2+(-1)^2}} j \\ \implies \dfrac{1}{\sqrt {2}} i+\dfrac{1}{\sqrt {2}} j \\ \implies \pm (\dfrac{1}{\sqrt {2}} i+\dfrac{1}{\sqrt {2}} j)$$