Differentiation of a function of three variables, dependent on a function of two variables. Chain rule 
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?
 A: 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}
