Implicit differentiation of a two variable function Let $f(u,v)$ be a differentiable function of two variables, and let z be a differentiable function of x and y defined implicitly by $f(xz,yz)$ = 0. Prove that $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -z$. I've got no idea how to start this. Any guidance would be appreciated. 
Thanks in advance!
 A: Define $g(x,y,z)=(xz,yz)$ and $h(x,y) = (x,y,z(x,y)).$  The equation $f(xz,yz)=0$ is equivalent to $(f\circ g \circ h)(x,y) = 0,$ where "$\circ$" denotes function composition.  Taking the total derivative of both sides of the composition equation yields
$$
  (Df)(Dg)(Dh) = 0
$$
where $Df,$ $Dg,$ and $Dh$ are the matrices of partial derivatives of the vector fields $f,$ $g,$ and $h.$
In matrix form, this is
$$
  (f_x \ f_y) 
  \left(
  \begin{array}{lll}
  z & 0 & x\\
  0 & z & y
  \end{array}
  \right)
  \left(
  \begin{array}{ll}
  1 & 0 \\
  0 & 1\\
  z_x & z_y
  \end{array}
  \right)
  = 0.
$$
Multiplying the last two matrices and transposing gives
$$
  \left(
  \begin{array}{ll}
  z+xz_x & yz_x \\
  xz_y & z+yz_y
  \end{array}
  \right)  
  \left(
  \begin{array}{l}
  f_x \\
  f_y
  \end{array}
  \right)
  =0.
$$  
If the matrix in this 2x2 linear system is non-singular, the solution
$(f_x, f_y)$ must be 0, meaning $f$ is constant, which is obviously not valid because then $f$ cannot be used to define $z.$  So the matrix must be singular, meaning the determinant $z^2 + z(xz_x + yz_y)=0,$ or $z+xz_x+yz_y=0.$
