Calculando o rotacional: use
\nabla\times\vec{F} = \bigl(\partial_{y}F_{3}-\partial_{z}F_{2}\bigr)\,\widehat{\imath} + \bigl(\partial_{z}F_{1}-\partial_{x}F_{3}\bigr)\,\widehat{\jmath} + \bigl(\partial_{x}F_{2}-\partial_{y}F_{1}\bigr)\,\widehat{k}.
Primeiro os parciais:
\partial_{y}F_{3} = \frac{\partial}{\partial y}\bigl(-y\,e^{x}\sin z\bigr) = -\,e^{x}\sin z, \quad \partial_{z}F_{2} = \frac{\partial}{\partial z}\bigl(x\,e^{y}\cos z\bigr) = -\,x\,e^{y}\sin z,
\partial_{z}F_{1} = \frac{\partial}{\partial z}\bigl(y^{2}+z^{2}\bigr) = 2\,z, \quad \partial_{x}F_{3} = \frac{\partial}{\partial x}\bigl(-y\,e^{x}\sin z\bigr) = -\,y\,e^{x}\sin z,
\partial_{x}F_{2} = \frac{\partial}{\partial x}\bigl(x\,e^{y}\cos z\bigr) = e^{y}\cos z, \quad \partial_{y}F_{1} = \frac{\partial}{\partial y}\bigl(y^{2}+z^{2}\bigr) = 2\,y.
Agora cada componente do rotacional:
(\nabla\times\vec{F})_{x} = (-\,e^{x}\sin z)\;-\;( -\,x\,e^{y}\sin z) = -\,e^{x}\sin z \;+\; x\,e^{y}\sin z,
(\nabla\times\vec{F})_{y} = 2\,z \;-\;( -\,y\,e^{x}\sin z) = 2\,z \;+\; y\,e^{x}\sin z,
(\nabla\times\vec{F})_{z} = e^{y}\cos z \;-\; 2\,y.
Logo
\nabla\times\vec{F} = \bigl(-\,e^{x}\sin z + x\,e^{y}\sin z\bigr)\widehat{\imath} + \bigl(2\,z + y\,e^{x}\sin z\bigr)\widehat{\jmath} + \bigl(e^{y}\cos z - 2\,y\bigr)\widehat{k}.