Answer
curl $F=-e^{-y} \cos z\hat{i}-e^{-z} \cos x\hat{j}-e^{-x} \cos y\hat{k}$;
div $F=-e^{-x} \sin y-e^{-y} \sin z-e^{-z} \sin x$
Work Step by Step
Let us consider $F=ai+bj+ck$
a) Since, curl $F=(\dfrac{\partial c}{\partial x}-\dfrac{\partial b}{\partial z}){i}+(\dfrac{\partial a}{\partial z}-\dfrac{\partial c}{\partial x}){j}+(\dfrac{\partial b}{\partial x}-\dfrac{\partial a}{\partial y}){k}$
Thus, curl $F=-e^{-y} \cos z\hat{i}-e^{-z} \cos x\hat{j}-e^{-x} \cos y\hat{k}$
b) $ div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
Thus,
div $F=\dfrac{\partial (e^{-x} \sin y)}{\partial x}+\dfrac{\partial (e^{-y} \sin z)}{\partial y}+\dfrac{\partial (e^{-z} \sin x)}{\partial z}$
Hence, div $F=-e^{-x} \sin y-e^{-y} \sin z-e^{-z} \sin x$
