Dual tensor for partial derivative, if it has any meaning I'm trying to find out some details about tensors, so my question maybe isn't quite correct.
What if $\omega$ is volume form in $(x,y,z)$ coordinates, then how to understand that $\frac{\partial}{\partial x}$ is a tensor and what is dual for that? Can I represent it as $\star\frac{\partial}{\partial x}=\frac{\partial}{\partial x}\star1=\frac{\partial}{\partial x}\omega$ or not? 
Please ask me if you need any clarification.
 A: The object $\tfrac{\partial}{\partial x}$ is understood here as a vector field (the coordinate vector field corresponding to the function $x \colon U \to \mathbb{R}$ for some open set $U \subseteq M$, where $M$ is the manifold in the question, but one can think of $\mathbb{R}^3$ for simplicity). Hence, $\tfrac{\partial}{\partial x}$ is a particular example of a $(1,0)$-tensor field. 

There is a notion of duality as in $V^*$ being the dual space for
  $V$, that is $V^*$ is the space of all linear functions on $V$. This
  works for any vector space $V$.

Now let us do some linear algebra. If $(V,g)$ is an inner product space, and $v,w \in V$ are some arbitrary vectors, we use the notation $\langle v, w \rangle$ instead of $g(v,w)$. 
Recall that tensors of type $(p,q)$ are elements of 
$$
\otimes^p V \otimes \otimes^q V^* =  \underbrace{V \otimes \dots \otimes V}_{p \text{ times}} \otimes \underbrace{V^* \otimes \dots \otimes V^*}_{q \text{ times}}
$$
Similarly, we can define tensor fields of type $p,q$ on $M$ as sections of tensor bundles $\bigotimes^p TM \otimes \bigotimes^q T^* M$. 
The inner product $g$ in $V$ canonically extends to tensors of type $(p,q)$. In particular, everything just said applies to exterior powers $\wedge^n V$ of $V$, and so on. We can speak then about the length of any tensor in this situation.
If the dimension of $V$ is $n$, the top exterior power $\wedge^n V$ of $V$ is $1$-dimensional. 
It looks like $\mathbb{R}$, but the direction is not chosen.
Using the inner product $g$ we can find an element $\omega$ of $\wedge^n V$ which has the length $1$. Notice that $\omega$ is a so-called $n$-vector in our approach. 
It is not unique, but choices only differ by the sign.
Fixing the choice introduces an orientation into $V$.
The chosen $\omega$ will be called the unit $n$-vector.
Let $\alpha$ be a $k$-vector, i.e. $\alpha \in \wedge^k V$. For any $(n-k)$-vector $\beta$ we have $\alpha \wedge \beta \in \wedge^n V$, and therefore $\alpha \wedge \beta = t \omega$ for some real number $t$.
Observe that for a fixed $\alpha$ the factor $t$ is a linear function on the vector space $\wedge^{n-k} V$, and therefore by the (finite-dimensional) Riesz theorem, there is a unique $\gamma$ such that 
$$
t(\beta) = g(\beta,\gamma) \tag{1}
$$
Definition. The Hodge dual of a $k$-vector $\alpha$ is defined as (n-k)-vector 
$$
\star \alpha = \gamma
$$
where $\gamma$ is as in $(1)$. (I am a little bit sloppy with the signature of $g$ here).
Thus we have 
$$
\alpha \wedge \beta = \langle \star \alpha, \beta \rangle \omega \tag{2}
$$
for all $\alpha \in \wedge^k V, \beta \in \wedge^{n-k} V$.

So, there are two notions of duality in the question! The way the OP
  stated the question, it sounds a bit ambiguous, but I gather that the
  Hodge duality is tacitly assumed, since the star operator $\star$ is
  used.

How to compute $\star \tfrac{\partial}{\partial x}$ in practice?  Some properties of the star operator may be very useful. I will use 
$$
\star \star = (-1)^{k(n-k)} \tag{3}
$$
without proof (it is easy to verify using bases). The equation $(2)$ can be rewritten as 
$$
\alpha \wedge \star \alpha = g(\alpha, \star \star \alpha) \omega =  (-1)^{k(n-k)} |\alpha|^2 \omega
$$
In the coordinate basis $(\tfrac{\partial}{\partial x},\tfrac{\partial}{\partial y},\tfrac{\partial}{\partial z})$ corresponding to the coordinate system $(x,y,z)$ given in the question, we have
$$
\tfrac{\partial}{\partial x} \wedge \star \tfrac{\partial}{\partial x} =  (-1)^{1 \cdot 2} \left| \tfrac{\partial}{\partial x} \right|^2 \frac{1}{\sqrt{\det(g)}} \tfrac{\partial}{\partial x} \wedge \tfrac{\partial}{\partial y} \wedge \tfrac{\partial}{\partial z}
$$
from where the answer may be simply read off:
$$
\star \tfrac{\partial}{\partial x} = \frac{\left| \tfrac{\partial}{\partial x} \right|^2}{\sqrt{\det(g)}} \tfrac{\partial}{\partial y} \wedge \tfrac{\partial}{\partial z}
$$

On $R^3$ with the standard metric and the coordinate system the answer
  simplifies to
  $$ \star \tfrac{\partial}{\partial x} =
 \tfrac{\partial}{\partial y} \wedge \tfrac{\partial}{\partial z} $$

A: Well, the partial derivative operator in your question is indeed (locally) a tensor, being a section of the tangent bundle. The dual one-form is $dx,$ i.e. it is the one-form defined by the following relationships: it evaluates to one on the vector field $\frac{\partial}{\partial x},$ and it evaluates to zero on the other basis vector fields $\frac{\partial}{\partial y}$ and $\frac{\partial}{\partial z}$. Then extend by linearity.
