Problem : Let E be the ellipsoid E = {(x, y, z) ∈ $\Bbb{R}^2$|$\left(\frac{x^2}{2}\right)$+$\left(\frac{y^2}{3}\right)$+ $z^2$ = 1}. Identify all points where E can be locally described as the graph of a function z = g(x, y).
The following is what I have on hand :
By definition of the Implicit Function Theorem,
Suppose f = $\left(\frac{x^2}{2}\right)$+$\left(\frac{y^2}{3}\right)$+ $z^2$ - 1 is a differentiable function with continuous function. Let ($x_o$, $y_o$, $z_o$) ∈ D, where D $\subset$ $\Bbb{R}^3$, such that f($x_o$, $y_o$, $z_o$)=0.
Suppose $\frac{\partial f}{\partial z}$($x_o$, $y_o$, $z_o$) $\neq$ 0
Then, there exists an open interval I, containing ($x_o$, $y_o$), and a differentiable function g : I $\to$ $\Bbb R$ with continuous derivative such that
1) g($x_o$, $y_o$) = $z_o$
2) f($x$, $y$, $z(x,y)$)=0 , for all $x, y$ $\in$ I
~at this point, I understand I would need to express the function z in terms of x and y~
$z$ = $\pm$$\sqrt{1-\left(\frac{x^2}{2}\right)-\left(\frac{y^2}{3}\right)}$
Differentiate :
w.r.t $x$,
$\frac{\partial f}{\partial x}$ $\cdot$ 1 + $\frac{\partial f}{\partial y}$ $\cdot$ 0 + $\frac{\partial f}{\partial z}$ $\cdot$ $\frac{\partial z}{\partial x}$ = 0
$\frac{\partial z}{\partial x}$ = - $\frac{f_x}{f_z}$ = - $\frac{x}{2z}$
w.r.t y,
$\frac{\partial f}{\partial y}$ $\cdot$ 0 + $\frac{\partial f}{\partial y}$ $\cdot$ 1 + $\frac{\partial f}{\partial z}$ $\cdot$ $\frac{\partial z}{\partial y}$ = 0
$\frac{\partial z}{\partial y}$ = - $\frac{f_y}{f_z}$ = - $\frac{y}{3z}$
$\Rightarrow$ $\frac{y}{x}$ = -$\frac{3x}{2y}$
From this point on, I rather confused as what I should do with the information on hand, what other information I require, and most importantly, how do I present my answer - specifically how do I find it and how do I know it is correct etc?
Im fairly new to this topic and I'm still trying to wrap my head around the concept of this theorem, hoping to get some advice and comment on my current workings (if they are even correct or useful), thanks in advance~