It is known, that antigradient $-\nabla f (x_0)$ is the direction of the steepest descent of the function $f(x)$ at point $x_0$. However, we can introduce another concept for choosing the best direction of function decreasing.

Given $f(x)$ and a point $x_0$. Define $B_\varepsilon(x_0) = \{x \in \mathbb{R}^n : d(x, x_0) = \varepsilon^2 \}$ as the set of points with distance $\varepsilon$ to $x_0$. Here we presume the existence of a distance function $d(x, x_0)$.

$x^* = \text{arg}\min_{x \in B_\varepsilon(x_0)} f(x)$

Than, we can define another steepest descent direction in terms of minimizer of function on a sphere:

$s = \lim_{\varepsilon \to 0} \frac{x^* - x_0}{\varepsilon}$

Let us assume that the distance is defined locally by some metric $A$:

$d(x, x_0) = (x-x_0)^\top A (x-x_0)$

Let us also consider first order Taylor approximation of a function $f(x)$ near the point $x_0$:

$\tag{A1} f(x_0 + \delta x) \approx f(x_0) + \nabla f(x_0)^\top \delta x$

Now we can explicitly pose a problem of finding $s$, as it was stated above.

$\begin{split} &\min_{\delta x \in \mathbb{R^n}} f(x_0 + \delta x) \\ \text{s.t.}\;& \delta x^\top A \delta x = \varepsilon^2 \end{split}$

Using $\text{(A1)}$ it can be written as:

$\begin{split} &\min_{\delta x \in \mathbb{R^n}} \nabla f(x_0)^\top \delta x \\ \text{s.t.}\;& \delta x^\top A \delta x = \varepsilon^2 \end{split}$

Using Lagrange multipliers method, we can easily conclude, that the answer is:

$\delta x = - \frac{2 \varepsilon^2}{\nabla f (x_0)^\top A^{-1} \nabla f (x_0)} A^{-1} \nabla f$

Which means, that new direction of steepest descent is nothing else, but $A^{-1} \nabla f(x_0)$.

Indeed, if the space is isotropic and $A = I$, we immediately have gradient descent formula, while Newton method uses local Hessian as a metric matrix.