Prove an identity using differential calculus to a problem connected to fluids 
Euler's equation for a incompressible inviscid fluid is 
$\displaystyle \frac{\partial \textbf{v}_t}{\partial t}+(\textbf{v}_t \cdot \nabla)\textbf{v}_t=-\nabla p_t$
where $\textbf{v}_t=\textbf{v}_t(r)$ is the fluid velocity and $p_t=p_t(r)$ is the pressure. Let $\nu_t(r)$ be the $1$-form associated to $\textbf{v}_t(r)$, ie,
$\displaystyle \nu_t=\sum_{i=1}^{3}v_t^idr^i$
Show that
$\displaystyle \frac{\partial \nu_t}{\partial t}+L_{\textbf{v}_t}\nu_t=-d(p_t-\frac{1}{2}v_t^2)$
where $v_t^2=\textbf{v}_t \cdot \textbf{v}_t$.
HINT: If $\omega=\omega_i dx^i$ is a $1$-form and $\mathbb{X}=X^ie_{(i)}$ a vector field on $\mathbb{R}^n$ then $\displaystyle L_{\mathbb{X}}\omega=(\frac{\partial X^j}{\partial x^i}\omega_j + X^j \frac{\partial \omega^i}{\partial x^j})dx^i$.

I am having serious problems attempting this question but I believe once going this question shouldnt be too bad. I believe it is ideal to start with the LHS and try and work to the RHS. 
I am unsure on how to express the first equation in a more differential form like form and so be able to use it in the question.
UPDATE 2: So I get so far
$$\begin{align} \frac{\partial \nu_t}{\partial t}+L_{\textbf{v}_t}\nu_t &= \frac{\partial \nu_t}{\partial t}+\left(\frac{\partial v^j}{\partial r^i}v_j + v^j \frac{\partial v_i}{\partial r^j}\right)dr^i.
\end{align} 
$$
working backwards,
$$ \begin{align} \left(-\frac{\partial p}{\partial r^i}+v^j\frac{\partial v^j}{\partial r^i} \right)dr^i &= \frac{\partial}{\partial r^i}(-p_t+\frac{1}{2}v^2)dr^i \\
&=  -d(p_t-\frac{1}{2}v^2)
\end{align}
$$ 
Somehow I need to connect the two parts.
Please answer if possible using basic differential calculus I am not familiar with using tensors, and Riemmannian manifolds, tangent spaces and bundles... I am only covering a basic course on differential calculus. Sorry for the inconvenience to those who looked at the question earlier. I believe my question should indicate the standard I am working at.
 A: Sketched derivation: Given a (time-independent) Riemannian manifold $(M,g)$. Let the connection $\nabla$ be the Levi-Civita connection. Let there be given a scalar function $p\in C^{\infty}(M)$; and a vector field $V\in\Gamma(TM)$. Define one-form/co-vector field $\nu:=V^{\flat}\in\Gamma(T^{\ast}M)$ via the musical isomorphism. 
The Euler's equation for a incompressible inviscid fluid reads in components
$$\tag{1} \partial_t V^k +\nabla^k p ~=~-(\nabla_VV)^k ~\equiv -V[V^k] -V^i \Gamma^k_{ij}V^j .$$
If we lower the vector index in eq. (1), we get
$$\tag{2}  \partial_t \nu_k +\partial_k p ~=~\frac{1}{2}V^iV^j\partial_k g_{ij}-V[\nu_k]. $$
Equation (2) leads to the sought-for equation
$$\tag{3} \partial_t \nu + \mathrm{d}p 
~=~ \frac{1}{2}V^iV^j\mathrm{d}g_{ij}-V[\nu_k] \mathrm{d}x^k 
~\stackrel{(4)}{\equiv}~\frac{1}{2}\mathrm{d}\left(\nu[V] \right)-{\cal L}_V\nu. $$
In the last step, we used the hint
$$\tag{4} {\cal L}_V\nu ~\equiv~V[\nu_k] \mathrm{d}x^k +\nu_k  \mathrm{d}V^k.$$
A: Use the hint with $\mathbb X = \mathbf v$ and $\omega = \nu$ (suppressing the $t$ subscript).  Then you get
$$\displaystyle L_{\mathbf{v}}\nu=\left(\frac{\partial v^j}{\partial r^i}v_j + v^j \frac{\partial v_i}{\partial r^j}\right)dr^i.$$
Now see that $$\frac{\partial v^j}{\partial r^i}v_j = \frac12 \frac{\partial }{\partial r^i}(v^j v_j) = \frac12 \frac{\partial }{\partial r^i}(v^2) ,$$ and $$v^j \frac{\partial v_i}{\partial r^j} = \mathbf v \cdot \nabla v_i.$$
It is mostly unwrapping the notation, and using implicit differentiation of $v^2$.
Remember $dr^i$ are simply the standard basis vectors expressed as $1$-forms.  And $dp$ is just a fancy way to write $\nabla p$.
