I'm asked in Ted Shifrin's Textbook Multivariable Mathematics to :
Suppose $h: \mathbb{R}^2 \to \mathbb{R}$ is $\mathcal{C^1}$ and $\frac{\partial h}{\partial x_2} \neq 0$. Show that the equation $h \begin{pmatrix}y/x \\z/x \end{pmatrix}=0$ defines $z$ $\\$ (locally) implicitly as a $\mathcal{C^1}$ function $z= \phi \begin{pmatrix}x \\y \end{pmatrix} $,
This is easy enough: I fix a $x \neq 0$, use the first assumption and then apply the implicit function Theorem.
However I'm also asked to show that $$x \ \frac{\partial \phi}{\partial x} + y \ \frac{\partial \phi}{\partial y} = \phi \begin{pmatrix}x \\y \end{pmatrix}$$
Which has me stuck. I thought about trying to show that the equation of the tangent plane at $\begin{pmatrix}x \\y \end{pmatrix}$ is zero at the point $\pmb 0$ but I can't go further than that.
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Edit after Ted Shifrin asked me to.
Since $h$ does change as we wiggle $x_2$ we can fix $x \neq 0$ to obtain a continuous one to one onto function $\mathbb{R} \to \mathbb{R}$, namely: $z/x$, and so we know that if we wiggle $z$ then $h \begin{pmatrix}y/x \\z/x \end{pmatrix}$ changes too (the partial derivative isn't zero) using the previous fact. This, together with the assumptions in the exercise about $h$, let us deduce that there is a $\mathcal{C^1}$ function $\phi \begin{pmatrix}x \\y \end{pmatrix} = z$ .
The equation of the tangent plane of the graph of $\phi$ at $\pmb x$ is:
$$g\begin{pmatrix}v \\ w \end{pmatrix}= \phi\begin{pmatrix}x \\ y \end{pmatrix}+ D\phi\begin{pmatrix}x \\ y \end{pmatrix}\begin{pmatrix}v-x\\ w-y \end{pmatrix}$$
So, evaluating $g$ at $\pmb 0$ we obtain:
$$g\begin{pmatrix}0 \\ 0 \end{pmatrix}= \phi\begin{pmatrix}x \\ y \end{pmatrix}+ D\phi\begin{pmatrix}x \\ y \end{pmatrix}\begin{pmatrix}-x\\ -y \end{pmatrix}$$
$$g\begin{pmatrix}0 \\ 0 \end{pmatrix}+ D\phi\begin{pmatrix}x \\ y \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix} = \phi\begin{pmatrix}x \\ y \end{pmatrix}$$
But I can't show how $g\begin{pmatrix}0 \\ 0 \end{pmatrix}=0$
Best Answer
Because $\dfrac{\partial h}{\partial x_2}\ne 0$, the Implicit Function Theorem tells us that locally we can express the level curve $h(x_1,x_2)=0$ as a graph $x_2 = F(x_1)$ for some smooth function $F$. Substituting $x_1=y/x$ and $x_2=z/x$, we get $\frac zx = F\big(\frac yx)$, so we have $$z=\phi(x,y)= xF(\frac yx),$$ with $F$ smooth and hence $\phi$ smooth. Now we just differentiate this equation (using product and chain rule), obtaining $$\frac{\partial\phi}{\partial x} = F(\tfrac yx) -\frac yx F'(\tfrac yx) \quad\text{and}\quad \frac{\partial\phi}{\partial y} = F'(\tfrac yx).$$ Now the rest is easy.
An alternative solution is to define $g(x,y,z)=(\frac yx,\frac zx)$ and set $H(x,y,z) = (h\circ g)(x,y,z)$. Now apply the chain rule to $H$ and use the implicit function theorem to show that since $\partial H/\partial z\ne 0$, we can locally express the level surface $H(x,y,z)=0$ as $z=\phi(x,y)$ for a smooth function $\phi$. This approach is good practice with the techniques.