How to prove that below quantity is a Third Rank Tensor $F^{ik}$ is an antisymmetric tensor. I want to prove that below quantity is a Third Rank Tensor.
$$\dfrac{\partial F_{ik}}{\partial x^{l}} + \dfrac{\partial F_{kl}}{\partial x^{i}} + \dfrac{\partial F_{li}}{\partial x^{k}}$$
I know the manipulations between contravarient and covariant tensors (using Metric Tensor). But How do I prove that something is a tensor with a specific rank?
This might seem trivial, but please, I am very new to this.
UPDATE:
Here is my attempt. I am transforming the below quantity to different coordinate system.
(I am going from i,k,l notation (unprimed) to j,t,f notation (primed).
$$F_{jt}' = \dfrac{\partial x^{i}}{\partial x'^{j}} \dfrac{\partial x^{k}}{\partial x'^{t}} F_{ik} $$
$$ \dfrac{\partial F_{jt}'}{\partial x'^{f}} = \dfrac{\partial}{\partial x^{l}} (\dfrac{\partial x^{i}}{\partial x'^{j}} \dfrac{\partial x^{k}}{\partial x'^{t}} F_{ik}) \dfrac{\partial x^{l}}{\partial x'^{f}}$$
Using the chain rule, I will get $\dfrac{\partial F_{ik}}{\partial x^{l}}$, which is one of the three terms in the given quantity; a d some other terms. Same procedure for other two terms and I will get the given quantity in addition with a whole lot of terms that wouldn't cancel out. How do I solve this? And how does the "antisymmetric" property help here? ($F^{ik}$ is antisymmetric, while in the equation, $F_{ik}$ is used).
 A: First notice that on account of the antisymmetry of $F_{ij}$ (notice that $F^{ij}$ will be antisymmetric whenever $F_{ij}$ is and viceversa) the object written
$$
T_{ijk}= \partial_i F_{jk} + \partial_{j}F_{ki} +\partial_{k}F_{ij}
$$
is totally antisymmetric under any index pair exchange (one can check this easily, such object is called the differential form $dF$, the exterior derivative of $F$). Now perform a coordinate change, $T_{ijk}$ will transform as
$$
T_{abc} = \frac{\partial x^i }{\partial x'^a}\frac{\partial x^j }{\partial x'^b}\frac{\partial x^k }{\partial x'^c}T_{ijk} + I_{abc}
$$
where this $I_{abc}$ is an inhomogeneous term given by:
$$
I_{abc} = \frac{\partial x^i}{\partial x'^a}\partial_i (\frac{\partial x^j}{\partial x'^b}\frac{\partial x^k}{\partial x'^c}) F_{jk} + \cdots
$$
such $I_{abc}$ will clearly be also totally antisymmetric under exchange of any pair of the indices $a,b,c$ (for it is defined as the difference of two objects that have this property each) Notice now that we can rewrite:
$$
I_{abc} = \frac{\partial }{\partial x'^a}(\frac{\partial x^j}{\partial x'^b}\frac{\partial x^k}{\partial x'^c}) F_{jk} + \cdots \\
=\frac{\partial^2 x^j}{\partial x'^a x'^b}\frac{\partial x^k}{\partial x'^c} F_{jk}  + \frac{\partial x^j}{\partial x'^b} \frac{\partial^2 x^j}{\partial x'^a x'^c} F_{jk} + \cdots  
$$
and they all vanish because the object is antisymmetric in the indices $a,b,c$ while the mixed partial derviatives are symmetric (remember that an object both symmetric and antisymmetric is zero), hence $T_{ijk}$ is a tensor.
