For $\displaystyle \Phi (x,y,z) = x^{xy} \sin(x+z) + y \ln (z) $
find the flux vector $\displaystyle \vec{J} = \triangledown \Phi$
and prove that curl $\displaystyle \vec{J} = 0$
Is this correct?
$\displaystyle \triangledown \Phi = \frac{\partial \Phi}{\partial x} i + \frac{\partial \Phi}{\partial y} j + \frac{\partial \Phi}{\partial z} k$
where
$\displaystyle \frac{\partial \Phi}{\partial x} = e^{xy} \cos(x+z) + \sin(x+z) y e^{xy}$
$\displaystyle \frac{\partial \Phi}{\partial y} = \sin(x+z) x e^{xy} + \ln(z)$
and
$\displaystyle \frac{\partial \Phi}{\partial z} = e^{xy} \cos(x+z) + \frac{y}{x} $
and
$\displaystyle \triangledown \Phi = (e^{xy} \cos(x+z) + \sin(x+z) y e^{xy}) i + (\sin(x+z) x e^{xy} + \ln(z) ) j + (e^{xy} \cos(x+z) + \frac{y}{x})k$
Is there anything more I need to do with this?