\[\begin{align} \hat U &= {\overrightarrow U \over |\overrightarrow U|} \\ &= {(0,-1) \over \sqrt{0^2+(-1)^2} } \\ &= {(0,-1) \over \sqrt{1} } \\ &= {(0,-1) \over 1 } \\ &= (0,-1) \,\,\,\,(I) \end{align}\]
\[\left\{ \begin{matrix} \begin{align} {\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)} &={\partial (x^2+y^2) \over \partial x}\Bigg|_{(x,y)=(1,1)} \\ {\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)} &={\partial (x^2+y^2) \over \partial y}\Bigg|_{(x,y)=(1,1)} \\ \end{align} \end{matrix} \right. \to\]
\[\left\{ \begin{matrix} \begin{align} {\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)} &=2x|_{(x,y)=(1,1)} \\ {\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)} &=2y|_{(x,y)=(1,1)} \\ \end{align} \end{matrix} \right. \to\]
\[\left\{ \begin{matrix} \begin{align} {\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)} &=2 \\ {\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)} &=2 \\ \end{align} \end{matrix} \right.\]
Portanto, o vetor gradiente é:
\[\begin{align} \nabla f(1,1) &= \Bigg({\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)}, {\partial f(x,y) \over \partial x}\Bigg|_{(x,y)=(1,1)}\Bigg) \\ &=(2,2) \,\,\,\,(II) \end{align}\]
\[\begin{align} D_Uf(x,y) &= \hat U \cdot \nabla f(1,1) \\ &=(0,-1) \cdot (2,2) \\ &=0\cdot 2-1\cdot 2 \\ &=0-2 \\ &=-2 \end{align}\]
Concluindo, pelos dados do enunciado, a derivada direcional é \(\boxed{D_Uf(x,y)=-2}\).
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