Implicit function theorem of ${ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=yf\left( \frac { z }{ y } \right) $

Question

If $z=z(x,y)$ This implicitly given by ${ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=yf\left( \frac { z }{ y } \right) $

I need to prove $({ x }^{ 2 }{ -y }^{ 2 }{ -z }^{ 2 })\frac { \partial z }{ \partial x } +2xy\frac { \partial z }{ \partial y } =2xz$

Well I tried to calculate $\frac { \partial z }{ \partial x }$ and $\frac { \partial z }{ \partial y }$ but i supposed that i´m doing wrong something.

Answer 1

As your question states, you have to use the implicit function theorem. One formulation of it as I recall is as follows:

Given $F(x,y,z)=0$, $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$, $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$, etc...

In your problem, you can write \begin{align*} F(x,y,z) = yf\left(\frac{z}{y}\right) - x^2-y^2-z^2. \end{align*} From there, you can see that \begin{align*} F_x &= -2x, \\ F_y &= f\left(\frac{z}{y}\right)+y\left(-\frac{z}{y^2}\right)f^\prime\left(\frac{z}{y}\right)-2y = f\left(\frac{z}{y}\right)-\frac{z}{y}f^\prime\left(\frac{z}{y}\right)-2y,\\ F_z &= f^\prime\left(\frac{z}{y}\right) - 2z. \end{align*} Using the implicit function theorem, we have that \begin{align*} \frac{\partial z}{\partial x} &= -\frac{F_x}{F_z} = -\frac{-2x}{f^\prime\left(\frac{z}{y}\right) - 2z}\\ \frac{\partial z}{\partial y} &= -\frac{F_y}{F_z} = -\frac{f\left(\frac{z}{y}\right)-\frac{z}{y}f^\prime\left(\frac{z}{y}\right)-2y}{f^\prime\left(\frac{z}{y}\right) - 2z}. \end{align*}

The right hand side of your expression (the one you have to show) becomes: \begin{align*} LHS &= \frac{2x(x^2-y^2-z^2)}{f^\prime\left(\frac{z}{y}\right) - 2z} -2xy\frac{f\left(\frac{z}{y}\right)-\frac{z}{y}f^\prime\left(\frac{z}{y}\right)-2y}{f^\prime\left(\frac{z}{y}\right) - 2z}\\ &= \frac{2x(x^2-y^2-z^2)-2x\overbrace{yf\left(\frac{z}{y}\right)}^{ x^2+y^2+z^2}+2xzf^\prime\left(\frac{z}{y}\right)+4xy^2}{f^\prime\left(\frac{z}{y}\right) - 2z}\\ &= \frac{2x(x^2-y^2-z^2)-2x(x^2+y^2+z^2)+2xzf^\prime\left(\frac{z}{y}\right)+4xy^2}{f^\prime\left(\frac{z}{y}\right) - 2z}\\ &= \frac{-4xy^2-4xz^2+2xzf^\prime\left(\frac{z}{y}\right)+4xy^2}{f^\prime\left(\frac{z}{y}\right) - 2z}\\ &=\frac{-4xz^2+2xzf^\prime\left(\frac{z}{y}\right)}{f^\prime\left(\frac{z}{y}\right) - 2z} = \frac{2xz\left(-2z+f^\prime\left(\frac{z}{y}\right)\right)}{f^\prime\left(\frac{z}{y}\right) - 2z}\\ LHS &= 2xz. \end{align*}

Which is what you needed to show. This problem required a bit of chain rule as well as some substitutions.