If $u_i(\mathbf{x})$ is a vector field, show that $\frac{\partial u_i}{\partial x_j}$ transforms as a second-rank tensor. 
If $u_i(\mathbf{x})$ is a vector field, show that $\frac{\partial
 u_i}{\partial x_j}$ transforms as a second-rank tensor.

I think is a very easy question I just don't fully understand what the question is asking me to show and so I think seeing a worked example would help me get my head around this a bit better. 
Thanks in advance for any help
 A: $\vec{u(\vec{x}})$ is a vector field , this means it can be written in terms of basis vectors and coordinates as : $u^{\nu}(\vec{x})\frac{\partial
}{\partial x^{\nu}}$ (using Einstein summation convention). The $\frac{\partial
}{\partial x^{\nu}}$ are the basis vectors.
The basis vectors $\frac{\partial
}{\partial x^{\nu}}$ transform under a general coordinate transformation
($x^{\lambda} \to x^{\lambda '}$) like a rank one tensor  : $\frac{\partial
}{\partial x^{\nu '}}=\frac{\partial
 x^{\lambda}}{\partial x^{\nu '}}\frac{\partial
}{\partial x^{\lambda}}$
It follows that $u^{\nu '}(\vec{x})\frac{\partial
}{\partial x^{\nu '}}= u^{\nu '}(\vec{x}) \frac{\partial
 x^{\lambda}}{\partial x^{\nu '}}\frac{\partial
}{\partial x^{\lambda}}=u^{\nu }(\vec{x})\frac{\partial
}{\partial x^{\nu }} \implies u^{\nu } \frac{\partial
 x^{\lambda '}}{\partial x^{\nu }} =u^{\lambda ' }$
Update, see comment @ContraKinta below:
$ \implies \frac{\partial
 u^{\lambda ' }}{\partial x^{\rho ' }} = \frac{\partial
}{\partial x^{\rho ' }}(u^{\nu }) \frac{\partial
 x^{\lambda '}}{\partial x^{\nu }} + 
u^{\nu }\frac{\partial
}{\partial x^{\rho ' }}( \frac{\partial
 x^{\lambda '}}{\partial x^{\nu }} )
= \frac{\partial
 u^{\nu }}{\partial x^{\mu }} \cdot \frac{\partial
 x^{\mu}}{\partial x^{\rho '  }}   \frac{\partial
 x^{\lambda '}}{\partial x^{\nu }}    + 
u^{\nu }\frac{\partial
 x^{\mu}}{\partial x^{\rho '}} \frac{\partial^2 x^{\lambda '}
 }{\partial x^{\mu } x^{\nu }}  $
Now if we demand that the transformation is an affine transformation we have that the second derivatives vanish : $\frac{\partial^2 x^{\lambda '}
}{\partial x^{\mu } x^{\nu }} = 0$.
That means that
$ \frac{\partial
 u^{\lambda ' }}{\partial x^{\rho ' }} 
= \frac{\partial
 u^{\nu }}{\partial x^{\mu }} \cdot \frac{\partial
 x^{\mu}}{\partial x^{\rho '  }}   \frac{\partial
 x^{\lambda '}}{\partial x^{\nu }} $ and the partials of the coordinates transform like second rank tensors.
