A question about The Implicit Function Theorem here is a problem im trying to solve for a few days, and Im not getting sucess. 
Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ such that $f\in C^1$ and $F(x,y,z)=f(y/x,z/x)$. Consider a level surface $S$ defined by $F(x,y,z)=0$ and $(x_0,y_0,z_0)\in S$. What are the conditions, in a neighborhood of $(x_0,y_0,z_0)$, that we can write $S$ in the form $z=g(x,y)$ ?
Also, show that at this conditions, its true that
$$x\frac{\partial g}{\partial x}(x,y)+y\frac{\partial g}{\partial y}(x,y)=g(x,y)$$
Well, here is what I've done:
To use the Implicit Function Theorem I choose $F(x_0,y_0,z_0)=0$ and I have to show that $\frac{\partial F}{\partial z}(x_0,y_0,z_0)\neq 0$. 
$$\frac{\partial F}{\partial z}(x_0,y_0,z_0)=\textrm{lim}_{t\rightarrow 0}\frac{F(x_0,y_0,z_0+t)-F(x_0,y_0,z_0)}{t}=\textrm{lim}_{t\rightarrow 0}\frac{F(x_0,y_0,z_0+t)}{t}=\textrm{lim}_{t\rightarrow 0}\frac{f(y_0/x_0,(z_0+t)/x_0)}{t}\neq 0$$
Here I tried something that I dont know if is right, and even if is right, i couldnt conclude the right answer from this.
$$\textrm{lim}_{t\rightarrow 0}\frac{f(y_0/x_0,(z_0+t)/x_0)}{t}=\textrm{lim}_{t\rightarrow 0}\frac{f(y_0/x_0,(z_0/x_0)+(t/x_0))}{t}=\frac{1}{x_0}\frac{\partial f}{\partial y}(y_0/x_0,z_0/x_0)$$
I did this variable change $u=t/x_0$ with $u\rightarrow 0$ and calculate the limit. But Im really not sure about it and I didnt found the last equality they are asking.
Thank you everyone one more time.
 A: Its possible to define $F:\mathbb{R}^3\rightarrow\mathbb{R}$ through the map $(x,y,z)\mapsto(y/x, z/x)\mapsto f(y/x, z/x)$. 
We can define $\alpha:\mathbb{R}^3\rightarrow\mathbb{R}^2$ such that $\alpha(x,y,z)=(y/x, z/x)$, this way we have $F=f\circ\alpha:\mathbb{R}^3\rightarrow\mathbb{R}$.
The derivate $(f\circ\alpha)'$ at the point $p=(x_0,y_0,z_0)$ is given by $(f\circ\alpha)'(p)=f'(\alpha(p))\cdot \alpha '(p)$, by the chain rule. We compute this:
$$\left( \begin{array}{cc}
\frac{\partial f}{\partial x}(\alpha(p)), & \frac{\partial f}{\partial y}(\alpha(p)) 
\end{array} \right)\cdot
\left( \begin{array}{ccc}
-\frac{y_0}{x_0^2}, & \frac{1}{x_0}, & 0\\
-\frac{z_0}{x_0^2}, & 0, & \frac{1}{x_0}
\end{array} \right)=\\ =
\left( \begin{array}{ccc}
-\frac{y_0}{x_0^2}\cdot\frac{\partial f}{\partial x}(\alpha(p)) -\frac{z_0}{x_0^2}\cdot\frac{\partial f}{\partial y}(\alpha(p)), \frac{1}{x_0}\cdot\frac{\partial f}{\partial x}(\alpha(p)), \frac{1}{x_0}\cdot\frac{\partial f}{\partial y}(\alpha(p))
\end{array} \right)
$$  
To apply the Implicit Function Theorem is necessary that $\frac{\partial F}{\partial z}(p)=\frac{1}{x_0}\cdot\frac{\partial f}{\partial y}(\alpha(p))$ be $\neq 0$, this way we need that $x_0\neq 0$ and $\frac{\partial f}{\partial y}(\alpha(p))\neq 0$.
And to finalize...
By the Implicit Function Theorem, we can write $z=g(x,y)$ in a neighborhood of $p=(x_0,y_0,z_0)$ such that $F(x,y,g(x,y))=0$ for all points in this neighborhood. Also, we have that 
$$\frac{\partial g}{\partial x}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}=\frac{y}{x}\cdot\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}+\frac{z}{x}$$
$$\frac{\partial g}{\partial y}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$
Therefore
$$x\cdot\frac{\partial g}{\partial x}+y\cdot\frac{\partial g}{\partial y}=x\cdot\Bigg( \frac{y}{x}\cdot\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}+\frac{z}{x}\Bigg)-y\cdot\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}=z=g(x,y)$$
