Assume I have a vector $v$ in $\mathbb{R}^3$ with elements $v=\left[x,y,z\right]^T$. Assume further that there is a plane in $\mathbb{R}^3$ for which $v$ is the steepest descent direction in $z$ if $z \ne 0$. Finally, assume that the coordinate system is rotated arbitrarily along the $z$ axis, so that the composition of the $xy$-components of $v$ is relatively arbitrary.
I wish to understand how the gradient of this plane $\left[\frac{dz}{dx},\frac{dz}{dy}\right]^T$ is composed. For this I would look at a few scenarios:
The $xy$ components of $v$ follow along either the $x$ or $y$ axis. In that case, the gradient would be $\left[\frac{z}{x},0\right]^T$ or $\left[0,\frac{z}{y}\right]^T$, depending on whether the $xy$ component was $\left[x,0\right]^T$ or $\left[0,y\right]^T$.
The $xy$ components of $v$ follow along a diagonal of the $xy$ plane. In that case, we would have $|x|=|y|$ and the gradient would be $\left[0.5\frac{z}{x},0.5\frac{z}{y}\right]^T$.
From how I see it, both cases could be explained with the following formula:
$$\left[\frac{dz}{dx},\frac{dz}{dy}\right]^T=\left[z\frac{|x|}{|x|+|y|},z\frac{|y|}{|x|+|y|}\right]^T$$
Now my question: Is this assumption correct? Does it generalize for arbitrary $xy$-components of $v$?
Solving the plane equation for $z$ to obtain $z$ as a function of $x$ and $y$, or, as I shall write it, $z = f(x,y)$, the gradient is, as you note, $\nabla f = \left[ \frac{\partial z}{\partial x},\frac{\partial z}{\partial y}\right]^T$. Its norm determines the slope in the direction of steepest ascent. So the slope in the direction of steepest descent is $-|\nabla f|$.
On the other hand, the slope determined by your vector $v = [x,y,z]^T$ is equal to its "rise" over its "run," or $\frac{z}{\sqrt{x^2+y^2}}$. Therefore \begin{equation} |\nabla f| = -\frac{z}{\sqrt{x^2+y^2}} \end{equation} Because $v$ points down the plane in the direction of steepest descent, $\lbrack x,y \rbrack^T$ must be parallel to $\nabla f$ and point in the opposite direction. So we could obtain the gradient by normalizing $-\lbrack x,y \rbrack^T$ and then multiplying by the length of the gradient. So \begin{equation} \nabla f = |\nabla f| \cdot \frac{-1}{\sqrt{x^2+y^2}} \left[ x,y \right]^T = \frac{z}{x^2+y^2} \left[ x,y \right]^T = \left[ \frac{zx}{x^2+y^2},\frac{zy}{x^2+y^2} \right]^T \end{equation} The short answer is that no, your formula isn't quite correct. You need $x^2 + y^2$ instead of $|x| + |y|$.