Question: Let $ f(s,t) $ be a differentiable function of two variables and let $h(x,y,z)=z\cdot f(\frac{x}{z}, \frac{y}{z})$. Simplify the expression $(x,y,z) \cdot \nabla h$
I am having trouble understanding how to go about this problem. My solution attempt: $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial s}\cdot \frac{\partial s}{\partial x} + \frac{\partial f}{\partial t}\cdot \frac{\partial t}{\partial x} = \frac{\partial f}{\partial s}\cdot\frac{1}{z}$
$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial s}\cdot \frac{\partial s}{\partial y} + \frac{\partial f}{\partial t}\cdot \frac{\partial t}{\partial y} = \frac{\partial f}{\partial s}\cdot\frac{1}{z}$
$\frac{\partial f}{\partial z} = \frac{\partial f}{\partial s}\cdot \frac{\partial s}{\partial z} + \frac{\partial f}{\partial t}\cdot \frac{\partial t}{\partial z} = \frac{\partial f}{\partial s}\cdot\frac{-x}{z^2} + \frac{\partial f}{\partial t}\cdot\frac{-y}{z^2}$
This should in turn give $\nabla h = (\frac{\partial f}{\partial s}, \frac{\partial f}{\partial t}, f(\frac{x}{z},\frac{y}{z}) + \frac{\partial f}{\partial s} \cdot \frac{-x}{z^2} + \frac{\partial f}{\partial t} \cdot \frac{-y}{z^2})$
And therefore $(x, y, z) \cdot \nabla h = x \cdot \frac{\partial f}{\partial s} + y \cdot \frac{\partial f}{\partial t} + z\cdot (f( \frac{x}{z} , \frac{y}{z} ) + \frac{\partial f}{\partial s} \cdot \frac{-x}{z^2} + \frac{\partial f}{\partial t} \cdot \frac{-y}{z^2}) = \frac{\partial f}{\partial s \cdot z} + \frac{\partial f}{\partial t \cdot z} + z \cdot f( \frac{x}{z}, \frac{y}{z})$
The answer is supposed to be $zf$, but I cannot figure out how to reach that conclusion. Could someone perhaps point me in the right direction; what am I doing wrong?
If $s = \frac{x}{z}$ and $t = \frac{y}{z}$ \begin{equation} \frac{\partial h}{\partial x} = z \frac{\partial f(s,t)}{\partial s} \frac{\partial s}{\partial x} = z \frac{\partial f(s,t)}{\partial s} \frac{1}{z} = \frac{\partial f(s,t)}{\partial s} \end{equation} \begin{equation} \frac{\partial h}{\partial y} = z \frac{\partial f(s,t)}{\partial t} \frac{\partial t}{\partial y} = z \frac{\partial f(s,t)}{\partial t} \frac{1}{z} = \frac{\partial f(s,t)}{\partial t} \end{equation} Using product rule first, \begin{equation} \frac{\partial h}{\partial z} = f(s,t) + z [\frac{\partial f(s,t)}{\partial z}] = f(s,t) + z [\frac{\partial f(s,t)}{\partial s} \frac{\partial s}{\partial z} + \frac{\partial f(s,t)}{\partial t} \frac{\partial t}{\partial z}] \end{equation} where $\frac{\partial s}{\partial z} = - \frac{x}{z^2} $ and $\frac{\partial t}{\partial z} = - \frac{y}{z^2} $. So \begin{equation} \frac{\partial h}{\partial z} = f(s,t) + z [-\frac{\partial f(s,t)}{\partial s} \frac{x}{z^2} - \frac{\partial f(s,t)}{\partial t} \frac{y}{z^2}] \end{equation} that is \begin{equation} \frac{\partial h}{\partial z} = f(s,t) - \frac{1}{z} [x\frac{\partial f(s,t)}{\partial s} + y\frac{\partial f(s,t)}{\partial t}] \end{equation} Now compute the inner product \begin{equation} \begin{bmatrix} x & y & z \end{bmatrix} \nabla h = x \frac{\partial h}{\partial x} + y \frac{\partial h}{\partial y} + z \frac{\partial h}{\partial z} \end{equation} Replacing \begin{equation} \begin{bmatrix} x & y & z \end{bmatrix} \nabla h = x \frac{\partial f(s,t)}{\partial s} + y \frac{\partial f(s,t)}{\partial t} + z \Big( f(s,t) - \frac{1}{z} [x\frac{\partial f(s,t)}{\partial s} + y\frac{\partial f(s,t)}{\partial t}] \Big) \end{equation} Expand \begin{equation} \begin{bmatrix} x & y & z \end{bmatrix} \nabla h = x \frac{\partial f(s,t)}{\partial s} + y \frac{\partial f(s,t)}{\partial t} + z f(s,t) -x\frac{\partial f(s,t)}{\partial s} - y\frac{\partial f(s,t)}{\partial t} \end{equation} which gives us the desired result \begin{equation} \begin{bmatrix} x & y & z \end{bmatrix} \nabla h = z f(s,t) \end{equation}