The notation 
\begin{equation} 
\frac{ \partial f_i}{ \partial x ^i }
\end{equation} 
means the diagonal elements of the matrix:
\begin{equation} 
J _{ ij} = \frac{ \partial f _i }{ \partial x ^j }
\end{equation} 
where $f_i$ is the component of the vector $\vec{f} (x)$.


----------

I found this very confusing a few weeks ago so. Here is the proof I wrote up for the identity based on the response I received to an earlier question of mine [here][1]:


Recall that if $ f (x) $ has one zero at $ x _0 $ then,
\begin{equation} 
\int d x \left| \frac{ df (x) }{ d x } \right|  _{ x = x _0  } \delta \left( f (x) \right) =   1
\end{equation}
 We want to generalize this to instead of having $ f (x) $ we have, $ {\mathbf{g}} ( {\mathbf{a}} ) $ for vectors of arbitrary size. To do this consider the Taylor expansion of $ {\mathbf{g}} $ around its root (we assume it only has one root, $ {\mathbf{a}} _0 $):
\begin{equation} 
g _i  ( {\mathbf{a}} ) = \overbrace{g _i ( {\mathbf{a}} _0 )}^0 + \sum _{ j} \frac{ \partial g _i }{ \partial a _j } \bigg|_{ a _0 } ( a _j - a _{ 0,j }) + ...
\end{equation} 
We want to insert this into a delta function, $ \delta ^{ ( n ) } ( {\mathbf{g}} ( {\mathbf{a}} ) ) $. This will only be nonzero near $ {\mathbf{a}} = {\mathbf{a}} _0 $. Thus we have,
\begin{align} 
\delta \left( {\mathbf{g}}  ( {\mathbf{a}} ) \right) & = \prod _i \delta \left( g _i ( {\mathbf{a}} ) \right) \\ 
&   =   \prod _i \delta \big( \sum _j J _{ ij} ( a _j - a _{ 0,j} )  \big)  
\end{align} 
where $ J _{ ij} $ is the Jacobian matrix defined by $ J _{ ij} \equiv \frac{ \partial g _{ i} }{  \partial a _j  } \big|_{ a _0 } $. We have,
\begin{align} 
\delta \left( {\mathbf{g}}  ( {\mathbf{a}} ) \right) & =   \delta \big( \sum _j J _{ 1j} ( a _j - a _{ 0,j} )  \big)  \delta \big( \sum _j J _{ 2j} ( a _j - a _{ 0,j} )  \big) ... 
\end{align} 
We now use the identity,
\begin{equation} 
\delta ( \alpha ( a - a _0 ) ) = \frac{ \delta ( a - a _0 ) }{ \left| \alpha \right| }
\end{equation} 
We choose to isolate each delta function in the equation above for a different $ a _j $:
\begin{align} 
\delta \big( {\mathbf{g}}  ( {\mathbf{a}} ) \big) & =   \frac{ \delta ( a _1 - a _{ 0,1 }  ) }{ \left| J _{ 1,1 } \right| }  \frac{ \delta ( a _2 - a _{ 0,2 }  ) }{ \left| J _{ 2,2 } \right| } ...
\end{align} 
If we take the Jacobian matrix to be greater then zero then we have the product:
\begin{equation} 
(  J _{ 1,1 } J _{ 2,2} .. ) ^{-1} = \frac{1}{ \det J } 
\end{equation} 
where we have used the fact that the determinant of $ J $ is independent of a unitary transformation. So we finally have,
\begin{align} 
\left( \int \prod _{ i} d a _i \right)  \delta ^{ ( n ) } \big( {\mathbf{g}}  ( {\mathbf{a}} ) \big) \det \big( \frac{ \partial g _i }{ \partial a _j  } \big)  & =   1
\end{align} 
where it is understood that the Jacobian matrix is evaluated at the root of $ {\mathbf{g}}  $.

We write the continuum generalization of this equation as,
\begin{equation} 
\int {\cal D} \alpha (x) \delta \left( G ( A ^\alpha ) \right) \det \left( \frac{ \delta G ( A ^\alpha ) }{ \delta \alpha } \right) = 1
\end{equation}  


  [1]: https://math.stackexchange.com/questions/672619/vector-delta-function-identity/674400#674400