Infinitesimal transformation of differential forms The infinitesimal differential forms $dx_1$ and $dx_2$ span a two-dimensional vector space (cotangent space). The transformation $f$ acts on $(x_1,x_2)$ like a coordinate transformation. Thus,
$$ dx_i \to du_i = \frac{\partial u_i}{\partial x_j}dx_j,\qquad i = 1,2.$$
Let's look at the infinitesimal transformation of $x_i$, i.e. $$ u_i(x) = x_i +\epsilon_i(x).$$ 
Show that if $f$ is a conformal transformation: 
$$ \omega(x)\delta_{ij} = -\frac{\partial \epsilon_j}{\partial x^i}-\frac{\partial \epsilon_i}{\partial x^j},$$
where $\omega(x) \in \mathbb{R}$ is a scale factor.
Hint: First, show that $$\delta_{ij}du^idu^j=\left(1+\omega(x)\right)\delta_{kl}dx^kdx^l.$$
I've already tried different things such as substituting the $du$'s in the expression given as a hint. All of this basically led to nothing, so it would be great if somebody could help me find the solution. How should I start?
Maybe I should mention that this is an exercise from a physics workbook, so this question has to be answered using the information given above. 
 A: You already have all the elements that you need. You just need to equate two expressions for the metric after a conformal transformation. Lets see it.


*

*First, its change under any transformation $x\mapsto u(x)$ can be calculated as
\begin{align}
  \delta_{ij}dx_idx_j\mapsto& \,\delta_{ij}du_idu_j =
  \delta_{ij}\frac{\partial u_i}{\partial x_k}
  \frac{\partial u_j}{\partial x_l}dx_kdx_l =
  \left(\delta_{ik}+\frac{\partial \epsilon_i}{\partial x_k}\right)
  \left(\delta_{jl}+\frac{\partial \epsilon_j}{\partial x_l}\right)
  \delta_{ij}dx_kdx_l\\
  =& \left(\delta_i^k\delta_j^l+
  \delta_{jl}\frac{\partial \epsilon_i}{\partial x_k}+
  \delta_{ik}\frac{\partial \epsilon_j}{\partial x_l}+
  O\left(\left(\frac{\partial\epsilon}{\partial x}
  \right)^2\right)\right)\delta_{ij}dx_kdx_l \\
  =& \left(\delta_{kl}+\frac{\partial\epsilon_l}{\partial x_k}
  +\frac{\partial\epsilon_k}{\partial x_l}+
  O\left(\left(\frac{\partial\epsilon}{\partial x}\right)^2\right)\right) dx_kdx_l
\end{align}

*On the other side, if it is a conformal transformation it should satisfy that after the transformation the metric is equal to
\begin{equation}
  \exp(-\omega(x))\delta_{ij}dx_idx_j=\left(1-
  \omega(x)+O\left(\omega(x)^2\right)\right)\delta_{ij}dx_idx_j.
\end{equation}
Now, neglecting quadratic or higher powers of $\omega$ and the derivatives of $\epsilon$ one gets:
\begin{equation}
  (1+\omega)\delta_{ij}dx_idx_j=\left(\delta_{kl}-
  \frac{\partial \epsilon_l}{\partial x_k}
  -\frac{\partial \epsilon_k}{\partial x_l}\right)dx_kdx_l
\end{equation}
And because the coefficient of each $dx_idx_j$ has to be equal on both sides:
\begin{equation}
  \omega\delta_{ij}=
  -\frac{\partial \epsilon_j}{\partial x_i}
  -\frac{\partial \epsilon_i}{\partial x_j}
\end{equation}
